Docsity
Log inSign up
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search Store documents

The best documents sold by students who completed their studies

Search through all study resources
Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

From our blog

Exams and Study
Go to the blog

Differential Equation Notes, Summaries of Mathematics

Indian Institute of ScienceMathematics
Prof. Anandamohan Ghosh

Very organised notes , once you read , you will know

Typology: Summaries

2022/2023

Uploaded on 05/03/2023

adrish556
adrish556 🇮🇳

2 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
G. NAGY – ODE January 26, 2012 11
Example 1.2.2:Find the solution of the problem given in Example 1.2.1 using the results
of Theorem 1.2.2.
Solution: We find the solution simply by using Eq. (1.2.3). First, find the integrating
factor function µas follows:
A(t)=−Zt
1
2
sds =−2£ln(t)−ln(1)§=−2 ln(t)⇒A(t)=ln(t−2).
The integrating factor is µ(t)=e−A(t), that is,
µ(t)=e−ln(t−2)=eln(t2)⇒µ(t)=t2.
Then, we compute the solution as follows:
y(t)= 1
t2h2+Z2
1
s24sds
i
=2
t2+1
t2Zt
1
4s3ds
=2
t2+1
t2(t4−1)
=2
t2+t2−1
t2⇒y(t)= 1
t2+t2.
C
1.2.2. The Bernoulli equation. Jacob Bernoulli found this equation in 1690 while he was
studying the isochrone problem: To find the curve such that the time taken by an object
sliding without friction in uniform gravity to its lowest point is independent of its starting
point. To solve this problem Jacob needed to solve a non-linear differential equation, which
is now called the Bernoulli equation. He did solve this equation in 1696, introducing a change
of unknown in such a way that the resulting differential equation for the new unknown is
linear. In what follows we explain this idea in more detail.
Definition 1.2.3. ABernoulli equation in the unknown function y, determined by the
functions p,q:(t1,t
2)→Rand a number n∈R,isthedifferential equation
y�=p(t)y+q(t)yn.(1.2.6)
In the case that n= 0 or n= 1 the Bernoulli equation reduces to a linear equation. The
interesting cases are when the Bernoulli equation is non-linear. The following result shows
how to find solutions to the non-linear Bernoulli equation.
Theorem 1.2.4 (Bernoulli).The function yis a solution of the Bernoulli equation
y�=p(t)y+q(t)yn,n�=1,
iffthe function v=1/y(n−1) is solution of the linear differential equation
v�=−(n−1)p(t)v−(n−1)q(t).
This result says, solve the linear equation for function v, a the compute the solution yof
the Bernoulli equation as y=(1/v)1/(n−1).
Proof of Theorem 1.2.4: Divide the Bernoulli equation by yn,
y�
yn=p(t)
yn−1+q(t).
Bernoulli equation:
G. NAGY – ODE January 26, 2012 11
Example 1.2.2:Find the solution of the problem given in Example 1.2.1 using the results
of Theorem 1.2.2.
Solution: We find the solution simply by using Eq. (1.2.3). First, find the integrating
factor function µas follows:
A(t)=−Zt
1
2
sds =−2£ln(t)−ln(1)§=−2 ln(t)⇒A(t)=ln(t−2).
The integrating factor is µ(t)=e−A(t), that is,
µ(t)=e−ln(t−2)=eln(t2)⇒µ(t)=t2.
Then, we compute the solution as follows:
y(t)= 1
t2h2+Z2
1
s24sds
i
=2
t2+1
t2Zt
1
4s3ds
=2
t2+1
t2(t4−1)
=2
t2+t2−1
t2⇒y(t)= 1
t2+t2.
C
1.2.2. The Bernoulli equation. Jacob Bernoulli found this equation in 1690 while he was
studying the isochrone problem: To find the curve such that the time taken by an object
sliding without friction in uniform gravity to its lowest point is independent of its starting
point. To solve this problem Jacob needed to solve a non-linear differential equation, which
is now called the Bernoulli equation. He did solve this equation in 1696, introducing a change
of unknown in such a way that the resulting differential equation for the new unknown is
linear. In what follows we explain this idea in more detail.
Definition 1.2.3. ABernoulli equation in the unknown function y, determined by the
functions p,q:(t1,t
2)→Rand a number n∈R,isthedifferential equation
y�=p(t)y+q(t)yn.(1.2.6)
In the case that n= 0 or n= 1 the Bernoulli equation reduces to a linear equation. The
interesting cases are when the Bernoulli equation is non-linear. The following result shows
how to find solutions to the non-linear Bernoulli equation.
Theorem 1.2.4 (Bernoulli).The function yis a solution of the Bernoulli equation
y�=p(t)y+q(t)yn,n�=1,
iffthe function v=1/y(n−1) is solution of the linear differential equation
v�=−(n−1)p(t)v−(n−1)q(t).
This result says, solve the linear equation for function v, a the compute the solution yof
the Bernoulli equation as y=(1/v)1/(n−1).
Proof of Theorem 1.2.4: Divide the Bernoulli equation by yn,
y�
yn=p(t)
yn−1+q(t).
12 G. NAGY – ODE january 26, 2012
Introduce the new unknown v=y−(n−1) and compute its derivative,
v�=£y−(n−1)§�=−(n−1)y−ny�⇒−
v�(t)
(n−1) =y�(t)
yn(t).
If we substitute vand this last equation into the Bernoulli equation we get
−v�
(n−1) =p(t)v+q(t)⇒v�=−(n−1)p(t)v−(n−1)q(t).
This establishes the Theorem. §
Example 1.2.3:Given any constants a0,b0, find every solution of the differential equation
y�=a0y+b0y3.
Solution: This is a Bernoulli equation. Divide the equation by y3,
y�
y3=a0
y2+b0.
Introduce the function v=1/y2and its derivative v�=−2(y�/y3), into the differential
equation above,
−v�
2=a0v+b0⇒v�=−2a0v−2b0⇒v�+2a0v=−2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t)=e2a0t,
°e2a0tv¢�=−2b0e2a0t⇒e2a0tv=−b0
a0
e2a0t+c
We obtain that v=ce
−2a0t−b0
a0
.Sincev=1/y2,
1
y2=ce
−2a0t−b0
a0⇒y(t)=±1
°ce
−2a0t−b0
a0¢1/2.
C
Example 1.2.4:Find every solution of the equation ty
�=3y+t5y1/3.
Solution: Rewrite the differential equation as
y�=3
ty+t4y1/3.
This is a Bernoulli equation for n=1/3. Divide the equation by y1/3,
y�
y1/3=3
ty2/3+t4.
Define the new unknown function v=1/y(n−1), that is, v=y2/3, compute is derivative,
v�=2
3
y�
y1/3, and introduce them in the differential equation,
3
2v�=3
tv+t4⇒v�−2
tv=2
3t4.
Integrate the linear equation of vusing the integrating factor method. The integrating
factor is computed as follows:
A(t)=Z2
tdt =2ln(t)=ln(t2),
12 G. NAGY – ODE january 26, 2012
Introduce the new unknown v=y−(n−1) and compute its derivative,
v�=£y−(n−1)§�=−(n−1)y−ny�⇒−
v�(t)
(n−1) =y�(t)
yn(t).
If we substitute vand this last equation into the Bernoulli equation we get
−v�
(n−1) =p(t)v+q(t)⇒v�=−(n−1)p(t)v−(n−1)q(t).
This establishes the Theorem. §
Example 1.2.3:Given any constants a0,b0, find every solution of the differential equation
y�=a0y+b0y3.
Solution: This is a Bernoulli equation. Divide the equation by y3,
y�
y3=a0
y2+b0.
Introduce the function v=1/y2and its derivative v�=−2(y�/y3), into the differential
equation above,
−v�
2=a0v+b0⇒v�=−2a0v−2b0⇒v�+2a0v=−2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t)=e2a0t,
°e2a0tv¢�=−2b0e2a0t⇒e2a0tv=−b0
a0
e2a0t+c
We obtain that v=ce
−2a0t−b0
a0
.Sincev=1/y2,
1
y2=ce
−2a0t−b0
a0⇒y(t)=±1
°ce
−2a0t−b0
a0¢1/2.
C
Example 1.2.4:Find every solution of the equation ty
�=3y+t5y1/3.
Solution: Rewrite the differential equation as
y�=3
ty+t4y1/3.
This is a Bernoulli equation for n=1/3. Divide the equation by y1/3,
y�
y1/3=3
ty2/3+t4.
Define the new unknown function v=1/y(n−1), that is, v=y2/3, compute is derivative,
v�=2
3
y�
y1/3, and introduce them in the differential equation,
3
2v�=3
tv+t4⇒v�−2
tv=2
3t4.
Integrate the linear equation of vusing the integrating factor method. The integrating
factor is computed as follows:
A(t)=Z2
tdt =2ln(t)=ln(t2),
12 G. NAGY – ODE january 26, 2012
Introduce the new unknown v=y−(n−1) and compute its derivative,
v�=£y−(n−1)§�=−(n−1)y−ny�⇒−
v�(t)
(n−1) =y�(t)
yn(t).
If we substitute vand this last equation into the Bernoulli equation we get
−v�
(n−1) =p(t)v+q(t)⇒v�=−(n−1)p(t)v−(n−1)q(t).
This establishes the Theorem. §
Example 1.2.3:Given any constants a0,b0, find every solution of the differential equation
y�=a0y+b0y3.
Solution: This is a Bernoulli equation. Divide the equation by y3,
y�
y3=a0
y2+b0.
Introduce the function v=1/y2and its derivative v�=−2(y�/y3), into the differential
equation above,
−v�
2=a0v+b0⇒v�=−2a0v−2b0⇒v�+2a0v=−2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t)=e2a0t,
°e2a0tv¢�=−2b0e2a0t⇒e2a0tv=−b0
a0
e2a0t+c
We obtain that v=ce
−2a0t−b0
a0
.Sincev=1/y2,
1
y2=ce
−2a0t−b0
a0⇒y(t)=±1
°ce
−2a0t−b0
a0¢1/2.
C
Example 1.2.4:Find every solution of the equation ty
�=3y+t5y1/3.
Solution: Rewrite the differential equation as
y�=3
ty+t4y1/3.
This is a Bernoulli equation for n=1/3. Divide the equation by y1/3,
y�
y1/3=3
ty2/3+t4.
Define the new unknown function v=1/y(n−1), that is, v=y2/3, compute is derivative,
v�=2
3
y�
y1/3, and introduce them in the differential equation,
3
2v�=3
tv+t4⇒v�−2
tv=2
3t4.
Integrate the linear equation of vusing the integrating factor method. The integrating
factor is computed as follows:
A(t)=Z2
tdt =2ln(t)=ln(t2),
Substituting in the Bernoulli equation
(linear first order ODE with variable coefficients)
Example: Solve
12 G. NAGY – ODE january 26, 2012
Introduce the new unknown v=y−(n−1) and compute its derivative,
v�=£y−(n−1)§�=−(n−1)y−ny�⇒−
v�(t)
(n−1) =y�(t)
yn(t).
If we substitute vand this last equation into the Bernoulli equation we get
−v�
(n−1) =p(t)v+q(t)⇒v�=−(n−1)p(t)v−(n−1)q(t).
This establishes the Theorem. §
Example 1.2.3:Given any constants a0,b0, find every solution of the differential equation
y�=a0y+b0y3.
Solution: This is a Bernoulli equation. Divide the equation by y3,
y�
y3=a0
y2+b0.
Introduce the function v=1/y2and its derivative v�=−2(y�/y3), into the differential
equation above,
−v�
2=a0v+b0⇒v�=−2a0v−2b0⇒v�+2a0v=−2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t)=e2a0t,
°e2a0tv¢�=−2b0e2a0t⇒e2a0tv=−b0
a0
e2a0t+c
We obtain that v=ce
−2a0t−b0
a0
.Sincev=1/y2,
1
y2=ce
−2a0t−b0
a0⇒y(t)=±1
°ce
−2a0t−b0
a0¢1/2.
C
Example 1.2.4:Find every solution of the equation ty
�=3y+t5y1/3.
Solution: Rewrite the differential equation as
y�=3
ty+t4y1/3.
This is a Bernoulli equation for n=1/3. Divide the equation by y1/3,
y�
y1/3=3
ty2/3+t4.
Define the new unknown function v=1/y(n−1), that is, v=y2/3, compute is derivative,
v�=2
3
y�
y1/3, and introduce them in the differential equation,
3
2v�=3
tv+t4⇒v�−2
tv=2
3t4.
Integrate the linear equation of vusing the integrating factor method. The integrating
factor is computed as follows:
A(t)=Z2
tdt =2ln(t)=ln(t2),
12 G. NAGY – ODE january 26, 2012
Introduce the new unknown v=y−(n−1) and compute its derivative,
v�=£y−(n−1)§�=−(n−1)y−ny�⇒−
v�(t)
(n−1) =y�(t)
yn(t).
If we substitute vand this last equation into the Bernoulli equation we get
−v�
(n−1) =p(t)v+q(t)⇒v�=−(n−1)p(t)v−(n−1)q(t).
This establishes the Theorem. §
Example 1.2.3:Given any constants a0,b0, find every solution of the differential equation
y�=a0y+b0y3.
Solution: This is a Bernoulli equation. Divide the equation by y3,
y�
y3=a0
y2+b0.
Introduce the function v=1/y2and its derivative v�=−2(y�/y3), into the differential
equation above,
−v�
2=a0v+b0⇒v�=−2a0v−2b0⇒v�+2a0v=−2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t)=e2a0t,
°e2a0tv¢�=−2b0e2a0t⇒e2a0tv=−b0
a0
e2a0t+c
We obtain that v=ce
−2a0t−b0
a0
.Sincev=1/y2,
1
y2=ce
−2a0t−b0
a0⇒y(t)=±1
°ce
−2a0t−b0
a0¢1/2.
C
Example 1.2.4:Find every solution of the equation ty
�=3y+t5y1/3.
Solution: Rewrite the differential equation as
y�=3
ty+t4y1/3.
This is a Bernoulli equation for n=1/3. Divide the equation by y1/3,
y�
y1/3=3
ty2/3+t4.
Define the new unknown function v=1/y(n−1), that is, v=y2/3, compute is derivative,
v�=2
3
y�
y1/3, and introduce them in the differential equation,
3
2v�=3
tv+t4⇒v�−2
tv=2
3t4.
Integrate the linear equation of vusing the integrating factor method. The integrating
factor is computed as follows:
A(t)=Z2
tdt =2ln(t)=ln(t2),
12 G. NAGY – ODE january 26, 2012
Introduce the new unknown v=y−(n−1) and compute its derivative,
v�=£y−(n−1)§�=−(n−1)y−ny�⇒−
v�(t)
(n−1) =y�(t)
yn(t).
If we substitute vand this last equation into the Bernoulli equation we get
−v�
(n−1) =p(t)v+q(t)⇒v�=−(n−1)p(t)v−(n−1)q(t).
This establishes the Theorem. §
Example 1.2.3:Given any constants a0,b0, find every solution of the differential equation
y�=a0y+b0y3.
Solution: This is a Bernoulli equation. Divide the equation by y3,
y�
y3=a0
y2+b0.
Introduce the function v=1/y2and its derivative v�=−2(y�/y3), into the differential
equation above,
−v�
2=a0v+b0⇒v�=−2a0v−2b0⇒v�+2a0v=−2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t)=e2a0t,
°e2a0tv¢�=−2b0e2a0t⇒e2a0tv=−b0
a0
e2a0t+c
We obtain that v=ce
−2a0t−b0
a0
.Sincev=1/y2,
1
y2=ce
−2a0t−b0
a0⇒y(t)=±1
°ce
−2a0t−b0
a0¢1/2.
C
Example 1.2.4:Find every solution of the equation ty
�=3y+t5y1/3.
Solution: Rewrite the differential equation as
y�=3
ty+t4y1/3.
This is a Bernoulli equation for n=1/3. Divide the equation by y1/3,
y�
y1/3=3
ty2/3+t4.
Define the new unknown function v=1/y(n−1), that is, v=y2/3, compute is derivative,
v�=2
3
y�
y1/3, and introduce them in the differential equation,
3
2v�=3
tv+t4⇒v�−2
tv=2
3t4.
Integrate the linear equation of vusing the integrating factor method. The integrating
factor is computed as follows:
A(t)=Z2
tdt =2ln(t)=ln(t2),
to get
12 G. NAGY – ODE january 26, 2012
Introduce the new unknown v=y−(n−1) and compute its derivative,
v�=£y−(n−1)§�=−(n−1)y−ny�⇒−
v�(t)
(n−1) =y�(t)
yn(t).
If we substitute vand this last equation into the Bernoulli equation we get
−v�
(n−1) =p(t)v+q(t)⇒v�=−(n−1)p(t)v−(n−1)q(t).
This establishes the Theorem. §
Example 1.2.3:Given any constants a0,b0, find every solution of the differential equation
y�=a0y+b0y3.
Solution: This is a Bernoulli equation. Divide the equation by y3,
y�
y3=a0
y2+b0.
Introduce the function v=1/y2and its derivative v�=−2(y�/y3), into the differential
equation above,
−v�
2=a0v+b0⇒v�=−2a0v−2b0⇒v�+2a0v=−2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t)=e2a0t,
°e2a0tv¢�=−2b0e2a0t⇒e2a0tv=−b0
a0
e2a0t+c
We obtain that v=ce
−2a0t−b0
a0
.Sincev=1/y2,
1
y2=ce
−2a0t−b0
a0⇒y(t)=±1
°ce
−2a0t−b0
a0¢1/2.
C
Example 1.2.4:Find every solution of the equation ty
�=3y+t5y1/3.
Solution: Rewrite the differential equation as
y�=3
ty+t4y1/3.
This is a Bernoulli equation for n=1/3. Divide the equation by y1/3,
y�
y1/3=3
ty2/3+t4.
Define the new unknown function v=1/y(n−1), that is, v=y2/3, compute is derivative,
v�=2
3
y�
y1/3, and introduce them in the differential equation,
3
2v�=3
tv+t4⇒v�−2
tv=2
3t4.
Integrate the linear equation of vusing the integrating factor method. The integrating
factor is computed as follows:
A(t)=Z2
tdt =2ln(t)=ln(t2),
to solve for v and finally
12 G. NAGY – ODE january 26, 2012
Introduce the new unknown v=y−(n−1) and compute its derivative,
v�=£y−(n−1)§�=−(n−1)y−ny�⇒−
v�(t)
(n−1) =y�(t)
yn(t).
If we substitute vand this last equation into the Bernoulli equation we get
−v�
(n−1) =p(t)v+q(t)⇒v�=−(n−1)p(t)v−(n−1)q(t).
This establishes the Theorem. §
Example 1.2.3:Given any constants a0,b0, find every solution of the differential equation
y�=a0y+b0y3.
Solution: This is a Bernoulli equation. Divide the equation by y3,
y�
y3=a0
y2+b0.
Introduce the function v=1/y2and its derivative v�=−2(y�/y3), into the differential
equation above,
−v�
2=a0v+b0⇒v�=−2a0v−2b0⇒v�+2a0v=−2b0.
The last equation is a linear differential equation for v. This equation can be solved using
the integrating factor method. Multiply the equation by µ(t)=e2a0t,
°e2a0tv¢�=−2b0e2a0t⇒e2a0tv=−b0
a0
e2a0t+c
We obtain that v=ce
−2a0t−b0
a0
.Sincev=1/y2,
1
y2=ce
−2a0t−b0
a0⇒y(t)=±1
°ce
−2a0t−b0
a0¢1/2.
C
Example 1.2.4:Find every solution of the equation ty
�=3y+t5y1/3.
Solution: Rewrite the differential equation as
y�=3
ty+t4y1/3.
This is a Bernoulli equation for n=1/3. Divide the equation by y1/3,
y�
y1/3=3
ty2/3+t4.
Define the new unknown function v=1/y(n−1), that is, v=y2/3, compute is derivative,
v�=2
3
y�
y1/3, and introduce them in the differential equation,
3
2v�=3
tv+t4⇒v�−2
tv=2
3t4.
Integrate the linear equation of vusing the integrating factor method. The integrating
factor is computed as follows:
A(t)=Z2
tdt =2ln(t)=ln(t2),
pf3
pf4
pf5

Partial preview of the text

Download Differential Equation Notes and more Summaries Mathematics in PDF only on Docsity!

Theorem 1.2.4 (Bernoulli). The function y is a solution of the Bernoulli equation

y

￿

= p(t) y + q(t) y

n

, n ￿= 1,

iff the function v = 1/y(n−1)^ is solution of the linear differential equation v

￿

= −(n − 1)p(t) v − (n − 1)q(t).

This result says, solve the linear equation for function v, a the compute the solution y of the Bernoulli equation as y = (1/v)^1 /(n−1). Proof of Theorem 1.2.4: Divide the Bernoulli equation by yn, y￿ yn^ = p(t) yn−^1

  • q(t).

Bernoulli equation : v ￿ = −(n − 1)p(t) v − (n − 1)q(t). says, solve the linear equation for function v, a the compute the solution y of quation as y = (1/v)^1 /(n−1). orem 1.2.4: Divide the Bernoulli equation by yn, y￿ yn^

p(t) yn−^1

  • q(t). 12 G. NAGY – ODE january 26, 2012

Introduce the new unknown v = y−(n−1)^ and compute its derivative,

v￿^ = £ y−(n−1) §￿ = −(n − 1)y−n^ y￿^ ⇒ − v￿(t) (n − 1) = y￿(t) yn(t) . If we substitute v and this last equation into the Bernoulli equation we get − v￿ (n − 1) = p(t) v + q(t) ⇒ v

￿

= −(n − 1)p(t) v − (n − 1)q(t).

This establishes the Theorem. § Example 1.2.3: Given any constants a 0 , b 0 , find every solution of the differential equation y￿^ = a 0 y + b 0 y^3. Solution: This is a Bernoulli equation. Divide the equation by y^3 , y￿ y^3 = a 0 y^2

  • b 0. Introduce the function v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential equation above, − v￿ 2 = a 0 v + b 0 ⇒ v

￿

= − 2 a 0 v − 2 b 0 ⇒ v

￿

+ 2a 0 v = − 2 b 0.

The last equation is a linear differential equation for v. This equation can be solved using

12 G. NAGY – ODE january 26, 2012 Introduce the new unknown v = y−(n−1)^ and compute its derivative, v￿^ =

y−(n−1)

= −(n − 1)y−n^ y￿^ ⇒ − v￿(t) (n − 1)

y￿(t) yn(t)

If we substitute v and this last equation into the Bernoulli equation we get

v￿ (n − 1) = p(t) v + q(t) ⇒ v ￿ = −(n − 1)p(t) v − (n − 1)q(t). This establishes the Theorem. § Example 1.2.3: Given any constants a 0 , b 0 , find every solution of the differential equation y￿^ = a 0 y + b 0 y^3. Solution: This is a Bernoulli equation. Divide the equation by y^3 , y￿ y^3

a 0 y^2

  • b 0. Introduce the function v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential equation above,

v￿ 2 = a 0 v + b 0 ⇒ v ￿ = − 2 a 0 v − 2 b 0 ⇒ v ￿

  • 2a 0 v = − 2 b 0. 12 G. NAGY – ODE january 26, 2012

Introduce the new unknown v = y−(n−1)^ and compute its derivative,

v

￿

=

£ y −(n−1)§￿^ = −(n − 1)y

−n

y

￿

⇒ −

v￿(t) (n − 1) = y￿(t) yn(t) . If we substitute v and this last equation into the Bernoulli equation we get − v￿ (n − 1) = p(t) v + q(t) ⇒ v

￿

= −(n − 1)p(t) v − (n − 1)q(t).

This establishes the Theorem. § Example 1.2.3: Given any constants a 0 , b 0 , find every solution of the differential equation y

￿

= a 0 y + b 0 y

3

.

Solution: This is a Bernoulli equation. Divide the equation by y^3 , y￿ y^3 = a 0 y^2

  • b 0. Introduce the function v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential equation above, − v￿ 2 = a 0 v + b 0 ⇒ v

￿

= − 2 a 0 v − 2 b 0 ⇒ v

￿

+ 2a 0 v = − 2 b 0.

The last equation is a linear differential equation for v. This equation can be solved using

2 a t Substituting in the Bernoulli equation (linear first order ODE with variable coefficients) Example: Solve

  • ODE january 26, 2012

e the new unknown v = y−(n−1)^ and compute its derivative,

v

￿

=

£ y −(n−1)§￿^ = −(n − 1)y

−n

y

￿

⇒ −

v￿(t) (n − 1) = y￿(t) yn(t) . titute v and this last equation into the Bernoulli equation we get − v￿ (n − 1) = p(t) v + q(t) ⇒ v￿^ = −(n − 1)p(t) v − (n − 1)q(t). lishes the Theorem. § e 1.2.3: Given any constants a 0 , b 0 , find every solution of the differential equation y

￿

= a 0 y + b 0 y

3

.

n: This is a Bernoulli equation. Divide the equation by y^3 , y￿ y^3 = a 0 y^2

  • b 0. e the function v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential above, − v￿ 2 = a 0 v + b 0 ⇒ v

￿

= − 2 a 0 v − 2 b 0 ⇒ v

￿

+ 2a 0 v = − 2 b 0.

quation is a linear di fferential equation for v. This equation can be solved using ating factor method. Multiply the equation by μ(t) = e^2 a^0 t,

12 G. NAGY – ODE january 26, 2012 Introduce the new unknown v = y−(n−1)^ and compute its derivative, v ￿ = £ y −(n−1)§￿^ = −(n − 1)y −n y ￿ ⇒ − v￿(t) (n − 1) = y￿(t) yn(t) . If we substitute v and this last equation into the Bernoulli equation we get − v￿ (n − 1) = p(t) v + q(t) ⇒ v￿^ = −(n − 1)p(t) v − (n − 1)q(t). This establishes the Theorem. § Example 1.2.3: Given any constants a 0 , b 0 , find every solution of the differential equation y￿^ = a 0 y + b 0 y^3. Solution: This is a Bernoulli equation. Divide the equation by y^3 , y￿ y^3 = a 0 y^2

  • b 0. Introduce the function v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential equation above, − v￿ 2 = a 0 v + b 0 ⇒ v￿^ = − 2 a 0 v − 2 b 0 ⇒ v￿^ + 2a 0 v = − 2 b 0. The last equation is a linear differential equation for v. This equation can be solved using the integrating factor method. Multiply the equation by μ(t) = e^2 a^0 t, 12 G. NAGY – ODE january 26, 2012 Introduce the new unknown v = y−(n−1)^ and compute its derivative, v￿^ = £ y−(n−1) §￿ = −(n − 1)y−n^ y￿^ ⇒ − v￿(t) (n − 1) = y￿(t) yn(t) . If we substitute v and this last equation into the Bernoulli equation we get − v￿ (n − 1) = p(t) v + q(t) ⇒ v￿^ = −(n − 1)p(t) v − (n − 1)q(t). This establishes the Theorem. § Example 1.2.3: Given any constants a 0 , b 0 , find every solution of the differential equation y￿^ = a 0 y + b 0 y^3. Solution: This is a Bernoulli equation. Divide the equation by y^3 , y￿ y^3 = a 0 y^2
  • b 0. Introduce the function v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential equation above, − v￿ 2 = a 0 v + b 0 ⇒ v￿^ = − 2 a 0 v − 2 b 0 ⇒ v￿^ + 2a 0 v = − 2 b 0. The last equation is a linear differential equation for v. This equation can be solved using the integrating factor method. Multiply the equation by μ(t) = e^2 a^0 t, 12 G. NAGY – ODE january 26, 2012 Introduce the new unknown v = y−(n−1)^ and compute its derivative, v￿^ = £ y−(n−1) §￿ = −(n − 1)y−n^ y￿^ ⇒ − v￿(t) (n − 1) = y￿(t) yn(t) . If we substitute v and this last equation into the Bernoulli equation we get − v￿ (n − 1) = p(t) v + q(t) ⇒ v￿^ = −(n − 1)p(t) v − (n − 1)q(t). This establishes the Theorem. § Example 1.2.3: Given any constants a 0 , b 0 , find every solution of the differential equation y￿^ = a 0 y + b 0 y^3. Solution: This is a Bernoulli equation. Divide the equation by y^3 , y￿ y^3 = a 0 y^2
  • b 0. Introduce the function v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential equation above, − v￿ 2 = a 0 v + b 0 ⇒ v￿^ = − 2 a 0 v − 2 b 0 ⇒ v￿^ + 2a 0 v = − 2 b 0. The last equation is a linear differential equation for v. This equation can be solved using the integrating factor method. Multiply the equation by μ(t) = e^2 a^0 t, ° e^2 a^0 tv ¢￿ = − 2 b 0 e^2 a^0 t^ ⇒ e^2 a^0 tv = − b 0 a 0 e^2 a^0 t^ + c to get january 26, 2012 known v = y−(n−1)^ and compute its derivative, £ y−(n−1) §￿ = −(n − 1)y−n^ y￿^ ⇒ − v￿(t) (n − 1) = y￿(t) yn(t) . and this last equation into the Bernoulli equation we get ￿

= p(t) v + q(t) ⇒ v￿^ = −(n − 1)p(t) v − (n − 1)q(t). Theorem. § iven any constants a 0 , b 0 , find every solution of the differential equation y￿^ = a 0 y + b 0 y^3. Bernoulli equation. Divide the equation by y^3 , y￿ y^3 = a 0 y^2

  • b 0. on v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential a 0 v + b 0 ⇒ v￿^ = − 2 a 0 v − 2 b 0 ⇒ v￿^ + 2a 0 v = − 2 b 0. linear di fferential equation for v. This equation can be solved using r method. Multiply the equation by μ(t) = e^2 a^0 t, ° e^2 a^0 tv ¢￿ = − 2 b 0 e^2 a^0 t^ ⇒ e^2 a^0 tv = − b 0 e^2 a^0 t^ + c to solve for v and finally Introduce the new unknown v = y−(n−1)^ and compute its derivative, v￿^ = £ y−(n−1) §￿ = −(n − 1)y−n^ y￿^ ⇒ − v￿(t) (n − 1) = y￿(t) yn(t) . If we substitute v and this last equation into the Bernoulli equation we get − v￿ (n − 1) = p(t) v + q(t) ⇒ v￿^ = −(n − 1)p(t) v − (n − 1)q(t). This establishes the Theorem. § Example 1.2.3: Given any constants a 0 , b 0 , find every solution of the differential equation y￿^ = a 0 y + b 0 y^3. Solution: This is a Bernoulli equation. Divide the equation by y^3 , y￿ y^3 = a 0 y^2
  • b 0. Introduce the function v = 1/y^2 and its derivative v￿^ = −2(y￿/y^3 ), into the differential equation above, − v￿ 2 = a 0 v + b 0 ⇒ v￿^ = − 2 a 0 v − 2 b 0 ⇒ v￿^ + 2a 0 v = − 2 b 0. The last equation is a linear differential equation for v. This equation can be solved using the integrating factor method. Multiply the equation by μ(t) = e^2 a^0 t, ° e^2 a^0 tv ¢￿ = − 2 b 0 e^2 a^0 t^ ⇒ e^2 a^0 tv = − b 0 a 0 e^2 a^0 t^ + c We obtain that v = c e−^2 a^0 t^ − b 0 a 0

. Since v = 1/y^2 ,

1 y^2 = c e−^2 a^0 t^ − b 0 a 0 ⇒ y(t) = ± 1 ° c e−^2 a^0 t^ − (^) ab^00 ¢ 1 / 2.

Separable equations

cated to solve than linear differential equations. However, it is known how to find solutions to particular types of non-linear differential equations. One of these types of equations is called separable differential equations.

Definition 1.3.1. Given functions h, g : R → R, a first order ODE on the unknown function y : R → R is called separable iff the ODE has the form

h(y) y￿(t) = g(t).

It is not difficult to see that a differential equation y￿(t) = f (t, y(t)) is separable iff

y￿^ =

g(t)

h(y)

⇔ f (t, y) =

g(t)

h(y)

.

Example 1.3.1:

(a) The differential equation

y￿(t) =

t^2

1 − y^2 (t)

is separable, since it is equivalent to

° 1 − y^2

¢ y￿(t) = t^2 ⇒

( g(t) = t

2 ,

h(y) = 1 − y^2.

(b) The differential equation y￿(t) + y^2 (t) cos(2t) = 0

is separable, since it is equivalent to

1

y^2

y￿(t) = − cos(2t) ⇒

8 <

:

g(t) = − cos(2t),

h(y) =

1

y^2

.

The functions g and h are not uniquely defined; another choice in this example is:

1

called separable differential equations.

Definition 1.3.1. Given functions h, g : R → R, a first order ODE on the unknown function

y : R → R is called separable iff the ODE has the form

h(y) y￿(t) = g(t).

It is not difficult to see that a differential equation y￿(t) = f (t, y(t)) is separable iff

y￿^ =

g(t)

h(y)

⇔ f (t, y) =

g(t)

h(y)

Example 1.3.1:

(a) The differential equation

y

￿

(t) =

t^2

1 − y^2 (t)

is separable, since it is equivalent to

1 − y

2 ¢^

y

￿

(t) = t

2

g(t) = t^2 ,

h(y) = 1 − y

2

(b) The differential equation

y￿(t) + y^2 (t) cos(2t) = 0

is separable, since it is equivalent to

y^2

y

￿

(t) = − cos(2t) ⇒

g(t) = − cos(2t),

h(y) =

y^2

The functions g and h are not uniquely defined; another choice in this example is:

g(t) = cos(2t), h(y) = −

y^2

(b) The differential equation

y￿(t) + y^2 (t) cos

is separable, since it is equivalent to

y^2

y￿(t) = − cos(2t) ⇒

The functions g and h are not uniquely defin

g(t) = cos(2t), h

(c) The linear differential equation y￿(t) = −a(t)

y

y

￿ (t) = −a(t) ⇒

(d) The constant coefficients linear differential eq since it is equivalent to

(−a 0 y + b 0 )

y￿(t) = 1 ⇒

(e) The differential equation y￿(t) = ey(t)^ + cos(t) (f) The linear differential equation y￿(t) = −a(t) separable.

Separable:

Remark: Theorem 1.1.2 says that Eq. (1.1.3) has infinitely ma each value of the constant c, which is not determined by the e Since the differential equation contains one derivative of the a solution of the differential equation requires to compute a integration introduces an integration constant. This constant o the constant c above. Proof of Theorem 1.1.2: The integrating factor method is Theorem 1.1.2. We first write down the differential equation a

y￿(t) − a y(t) = b,

and then multiply the differential equation by the exponentia is the constant coefficient a in the differential equation multip function is an integrating factor for the differential equation. particular function, e−at, is explained in Lemma 1.1.3, below. £ y￿(t) − a y(t)

§ e−at^ = b e−at^ ⇔ e−at^ y￿(t) − a e

This exponential is chosen because of the following property,

−a e−at^ =

° e−ay^

¢￿ .

Introducing this property into the differential equation we get

e−at^ y￿(t) +

° e−at

¢￿ y(t) = b e−at.

Now the product rule for derivatives implies that the left-hand si £ e−at^ y(t)

§￿ = b e−at.

Not separable:

16 G. NAGY – ODE january 26, 2012

We see in the examples above that a linear ODE with function b = 0 is separable, so the solutions of such equation can be computed using both the result we show below and the integrating factor method studied in Sect. 1.2. We now describe a way to find solutions to every first order separable differential equation.

Theorem 1.3.2 (Separable equations). If the functions g, h : R → R are continuous, with h ￿= 0 and with primitives G and H, respectively; that is,

G

￿ (t) = g(t), H

￿ (u) = h(u),

then, the separable ODE h(y) y￿^ = g(t) (1.3.1)

has infinitely many solutions y : R → R satisfying the algebraic equation

H(y(t)) = G(t) + c, (1.3.2)

where c ∈ R is arbitrary.

Before presenting the proof of the Theorem above we use it in the example below and we explicitly show how to find the primitives G and H of the given functions g and h.

Example 1.3.2: Find all solutions y : R → R to the ODE

y￿(t) =

t^2

1 − y^2 (t)

Solution: Theorem 1.3.2 tell us how to obtain the solution y. Writing Eq. (1.3.3) as ° 1 − y^2

y￿(t) = t^2 ,

If g & h are continuous functions we can define the primitives G & H respectively such that

G. NAGY – ODE January 26, 2012 17

Sometimes is difficult to find the inverse of function H. This is the case in Example 1.3.2. In such cases we leave the solution y written in implicit form. If H−^1 is simple to compute, we express the solution y in explicit form. We now show the proof of Theorem 1.3.2, which is based in an integration by substitution. Proof of Theorem 1.3.2: Integrate on both sides in (1.3.1),

h(y(t)) y￿(t) = g(t) ⇒

Z

h(y(t)) y￿(t) dt =

Z

g(t) dt + c,

where c is an arbitrary constant. Introduce on the left-hand side of the second equation above the substitution

u = y(t), du = y￿(t) dt,

which implies, Z

h(y(t)) y￿(t) dt =

Z

h(u)du ⇒

Z

h(u) du =

Z

g(t) dt + c.

G. NAGY – ODE January 26, 2012 17

es is difficult to find the inverse of function H. This is the case in Example 1.3.2. s we leave the solution y written in implicit form. If H−^1 is simple to compute, he solution y in explicit form. We now show the proof of Theorem 1.3.2, which

n integration by substitution.

heorem 1.3.2: Integrate on both sides in (1.3.1),

h(y(t)) y

￿ (t) = g(t) ⇒

Z

h(y(t)) y

￿ (t) dt =

Z

g(t) dt + c,

an arbitrary constant. Introduce on the left-hand side of the second equation bstitution

u = y(t), du = y

￿ (t) dt,

ies,

G. NAGY – ODE January 26, 2012 17

Sometimes is difficult to find the inverse of function H. This is the case in Example 1.3.2. In such cases we leave the solution y written in implicit form. If H−^1 is simple to compute, we express the solution y in explicit form. We now show the proof of Theorem 1.3.2, which is based in an integration by substitution. Proof of Theorem 1.3.2: Integrate on both sides in (1.3.1),

h(y(t)) y￿(t) = g(t) ⇒

Z

h(y(t)) y￿(t) dt =

Z

g(t) dt + c,

where c is an arbitrary constant. Introduce on the left-hand side of the second equation above the substitution

u = y(t), du = y￿(t) dt,

which implies, Z

h(y(t)) y￿(t) dt =

Z

h(u)du ⇒

Z

h(u) du =

Z

g(t) dt + c.

Recalling that H and G are the primitives of h and g, respectively,

G. NAGY – ODE January 26, 2012 17

Sometimes is difficult to find the inverse of function H. This is the case in Example 1.3.2. In such cases we leave the solution y written in implicit form. If H−^1 is simple to compute, we express the solution y in explicit form. We now show the proof of Theorem 1.3.2, which is based in an integration by substitution. Proof of Theorem 1.3.2: Integrate on both sides in (1.3.1),

h(y(t)) y￿(t) = g(t) ⇒

Z

h(y(t)) y￿(t) dt =

Z

g(t) dt + c,

where c is an arbitrary constant. Introduce on the left-hand side of the second equation above the substitution

u = y(t), du = y￿(t) dt,

which implies, Z

h(y(t)) y￿(t) dt =

Z

h(u)du ⇒

Z

h(u) du =

Z

g(t) dt + c.

Recalling that H and G are the primitives of h and g, respectively,

H(u) =

Z

h(u) du, G(t) =

Z

g(t) dt,

Substitute to get

G. NAGY – ODE January 26, 2012 17

Sometimes is difficult to find the inverse of function H. This is the case in Example 1.3.2. In such cases we leave the solution y written in implicit form. If H−^1 is simple to compute, we express the solution y in explicit form. We now show the proof of Theorem 1.3.2, which is based in an integration by substitution. Proof of Theorem 1.3.2: Integrate on both sides in (1.3.1),

h(y(t)) y￿(t) = g(t) ⇒

Z

h(y(t)) y￿(t) dt =

Z

g(t) dt + c,

where c is an arbitrary constant. Introduce on the left-hand side of the second equation above the substitution

u = y(t), du = y￿(t) dt,

which implies, Z

h(y(t)) y￿(t) dt =

Z

h(u)du ⇒

Z

h(u) du =

Z

g(t) dt + c.

Recalling that H and G are the primitives of h and g, respectively,

H(u) =

Z

h(u) du, G(t) =

Z

g(t) dt,

then we obtain that

H(u) = G(t) + c.

Substitute y(t) back in the place of u. We conclude that the solution function y satisfies the algebraic equation ° ¢

G. NAGY – ODE January 26, 2012 17

Sometimes is difficult to find the inverse of function H. This is the case in Example 1.3.2. In such cases we leave the solution y written in implicit form. If H−^1 is simple to compute, we express the solution y in explicit form. We now show the proof of Theorem 1.3.2, which is based in an integration by substitution. Proof of Theorem 1.3.2: Integrate on both sides in (1.3.1),

h(y(t)) y￿(t) = g(t) ⇒

Z

h(y(t)) y￿(t) dt =

Z

g(t) dt + c,

where c is an arbitrary constant. Introduce on the left-hand side of the second equation above the substitution

u = y(t), du = y￿(t) dt,

which implies, Z

h(y(t)) y￿(t) dt =

Z

h(u)du ⇒

Z

h(u) du =

Z

g(t) dt + c.

Recalling that H and G are the primitives of h and g, respectively,

H(u) =

Z

h(u) du, G(t) =

Z

g(t) dt,

then we obtain that

H(u) = G(t) + c.

Substitute y(t) back in the place of u. We conclude that the solution function y satisfies the algebraic equation

H

° y(t)

¢ = G(t) + c.

Homogeneous equations 1.3.2. Homogeneous equations. Sometimes a differential equation is not separable but a change of unknown in the equation can transform a non-separable equation into a separable equation. This is the case for a type of differential equations called homogeneous equations. Definition 1.3.4. The first order differential equation of the form y￿(t) = f ° t, y(t) ¢ is called homogeneous iff for every real numbers t, u and every c ￿= 0 the function f satisfies f (ct, cu) = f (t, u). A differential equation is homogeneous when the function f is invariant under the change of scale of its arguments. It is not difficult to prove that a function with values f (t, u) having the property above must depend on (t, u) only through their quotient. That is, there exists a function F : R → R such that f (t, u) = F ≥ u t ¥ . Therefore, a first order differential equation is homogeneous iff it has the form y￿(t) = F ≥ y(t) t ¥ . Example 1.3.8: Show that the equation below is homogeneous, (t − y) y￿^ − 2 y + 3t + y^2 t = 0. Solution: Rewrite the equation in the standard form (t − y) y￿^ = 2y − 3 t − y^2 t ⇒ y￿^ = ≥ 2 y − 3 t − y^2 t ¥ (t − y) . Divide numerator and denominator by t. We get, y￿^ = ≥ 2 y − 3 t − y^2 t ¥ (t − y) ≥ (^1) t ¥ ≥ 1 t ¥ ⇒^ y ￿ (^) = 2 ≥ (^) y t ¥ − 3 − ≥ (^) y t ¥ 2 h 1 − ≥ y t ¥i (^). A differential equation is homogeneous when the function f is inva of scale of its arguments. It is not difficult to prove that a function wi the property above must depend on (t, u) only through their quotien a function F : R → R such that f (t, u) = F ≥ u t ¥ . Therefore, a first order differential equation is homogeneous iff it ha y￿(t) = F ≥ y(t) t ¥ . Example 1.3.8: Show that the equation below is homogeneous, (t − y) y￿^ − 2 y + 3t + y^2 t = 0. Solution: Rewrite the equation in the standard form (t − y) y￿^ = 2y − 3 t − y^2 t ⇒ y￿^ = ≥ 2 y − 3 t − y (t − y) Divide numerator and denominator by t. We get, y￿^ = ≥ 2 y − 3 t − y^2 t ¥ (t − y) ≥ 1 t ¥ ≥ (^1) t ¥ ⇒^ y ￿ (^) = 2 ≥ y t ¥ − 3 − h 1 − ≥ (^) y t ¥ We conclude that the differential equation is homogeneous, because the equation above depends only on y/t. Indeed, we have shown where f (t, y) = 2 y − 3 t − (y^2 /t) t − y , F (x) = 2 x − 3 − 1 − x

A differential equation is homogeneous when the function f is invar of scale of its arguments. It is not difficult to prove that a function wit the property above must depend on (t, u) only through their quotient a function F : R → R such that f (t, u) = F ≥ (^) u t ¥ . Therefore, a first order differential equation is homogeneous iff it has y￿(t) = F ≥ (^) y(t) t ¥ . Example 1.3.8: Show that the equation below is homogeneous, (t − y) y￿^ − 2 y + 3t + y^2 t = 0. Solution: Rewrite the equation in the standard form (t − y) y￿^ = 2y − 3 t − y^2 t ⇒ y￿^ = ≥ 2 y − 3 t − y t (t − y) Divide numerator and denominator by t. We get, y￿^ = ≥ 2 y − 3 t − y^2 t ¥ (t − y) ≥ (^1) t ¥ ≥ (^1) t ¥ ⇒^ y￿^ = 2 ≥ (^) y t ¥ − 3 − ≥ h 1 − ≥ (^) y t ¥ We conclude that the differential equation is homogeneous, because the equation above depends only on y/t. Indeed, we have shown t where f (t, y) = 2 y − 3 t − (y^2 /t) t − y , F (x) = 2 x − 3 − x 1 − x G. NAGY – ODE January 26, 2012 21 C : Determine whether the equation below is homogeneous, y￿^ = t^2 1 − y^3 . iding numerator and denominator by t^3 we obtain y￿^ = t^2 (1 − y^3 ) ≥ (^1) t^3 ¥ ≥ (^1) t^3 ¥ ⇒^ y ￿ (^) = ≥ (^1) t ¥ ≥ (^1) t^3 ¥ − ≥ (^) y t ¥ 3. t the di fferential equation is not homogeneous. C (Homogeneous equations). If the differential equation y￿(t) = f ° t, y(t) ¢ then the di fferential equation for the unknown v(t) = y(t) t is separable. ys that homogeneous equations can be transformed into separable equations. al equation for y is homogeneous, then the differential equation for v = y/t refore, an homogeneous equation for y is solved as follows: First, find the on for v; second, solve for v in implicit or explicit form; third, replace back Examples G. NAGY – ODE January 26, 2012 21

C

Example 1.3.9: Determine whether the equation below is homogeneous, y ￿ = t^2 1 − y^3

Solution: Dividing numerator and denominator by t^3 we obtain y￿^ = t^2 (1 − y^3 )

t^3

t^3 ¥ ⇒^ y

t

t^3

y t

We conclude that the differential equation is not homogeneous. C Theorem 1.3.5 (Homogeneous equations). If the differential equation y￿(t) = f

t, y(t)

is homogeneous, then the differential equation for the unknown v(t) = y(t) t is separable. The Theorem says that homogeneous equations can be transformed into separable equations. If the differential equation for y is homogeneous, then the differential equation for v = y/t is separable. Therefore, an homogeneous equation for y is solved as follows: First, find the separable equation for v; second, solve for v in implicit or explicit form; third, replace back y = t v. ⟹ not homogeneous operty above must depend on ( t, u) only through their quotient. That is, there exists tion F : R → R such that f (t, u) = F

u t

efore, a first order differential equation is homogeneous iff it has the form y￿(t) = F

y(t) t

ple 1.3.8: Show that the equation below is homogeneous, (t − y) y￿^ − 2 y + 3t + y^2 t

tion: Rewrite the equation in the standard form (t − y) y￿^ = 2y − 3 t − y^2 t ⇒ y￿^ =

2 y − 3 t − y^2 t

(t − y)

e numerator and denominator by t. We get, y￿^ =

2 y − 3 t − y^2 t

(t − y)

t

t ¥ ⇒^ y

y t

y t

h 1 −

y t ¥i (^). nclude that the di fferential equation is homogeneous, because the right-hand side of uation above depends only on y/t. Indeed, we have shown that f (t, y) = F (y/t), e f (t, y) = 2 y − 3 t − (y^2 /t) t − y , F (x) = 2 x − 3 − x^2 1 − x

y￿(t) = F ≥ (^) y(t) t ¥ . Example 1.3.8: Show that the equation below is homogeneous, (t − y) y￿^ − 2 y + 3t + y^2 t = 0. Solution: Rewrite the equation in the standard form (t − y) y￿^ = 2y − 3 t − y^2 t ⇒ y￿^ = ≥ 2 y − 3 t − y^2 t ¥ (t − y) . Divide numerator and denominator by t. We get, y￿^ = ≥ 2 y − 3 t − y^2 t ¥ (t − y) ≥ 1 t ¥ ≥ 1 t ¥ ⇒^ y ￿ (^) = 2 ≥ y t ¥ − 3 − ≥ y t ¥ 2 h 1 − ≥ (^) y t ¥i (^). We conclude that the differential equation is homogeneous, because the right-hand side of the equation above depends only on y/t. Indeed, we have shown that f (t, y) = F (y/t), where f (t, y) = 2 y − 3 t − (y^2 /t) t − y , F (x) = 2 x − 3 − x^2 1 − x , t Example 1.3.8: Show that the equation below is homogeneous, (t − y) y￿^ − 2 y + 3t + y^2 t = 0. Solution: Rewrite the equation in the standard form (t − y) y￿^ = 2y − 3 t − y^2 t ⇒ y￿^ = ≥ 2 y − 3 t − y^2 t ¥ (t − y) . Divide numerator and denominator by t. We get, y￿^ = ≥ 2 y − 3 t − y^2 t ¥ (t − y) ≥ (^1) t ¥ ≥ (^1) t ¥ ⇒^ y ￿ (^) = 2 ≥ (^) y t ¥ − 3 − ≥ (^) y t ¥ 2 h 1 − ≥ (^) y t ¥i (^). We conclude that the differential equation is homogeneous, because the right-hand side of the equation above depends only on y/t. Indeed, we have shown that f (t, y) = F (y/t), where f (t, y) = 2 y − 3 t − (y^2 /t) t − y , F (x) = 2 x − 3 − x^2 1 − x , ⟹ (^) homogeneous

Example 1.3.9: Determine whether the equation below is homogeneous,

y

￿

=

t^2 1 − y^3 . Solution: Dividing numerator and denominator by t^3 we obtain y

￿

=

t^2 (1 − y^3 ) ≥ 1 t^3 ¥ ≥ 1 t^3 ¥ ⇒^ y

￿

=

≥ 1 t ¥ ≥ 1 t^3 ¥ − ≥ y t ¥ 3. We conclude that the differential equation is not homogeneous. C Theorem 1.3.5 (Homogeneous equations). If the differential equation y￿(t) = f ° t, y(t) ¢ is homogeneous, then the differential equation for the unknown v(t) = y(t) t is separable. The Theorem says that homogeneous equations can be transformed into separable equations. If the differential equation for y is homogeneous, then the differential equation for v = y/t is separable. Therefore, an homogeneous equation for y is solved as follows: First, find the separable equation for v; second, solve for v in implicit or explicit form; third, replace back y = t v. Proof of Theorem 1.3.5: If y￿^ = f (t, y) is homogeneous, then it can be written as y￿^ = F (y/t) for some function F. Introduce v = y/t. This means, y(t) = t v(t) ⇒ y

￿

(t) = v(t) + t v

￿

(t).

Introducing these expressions into the differential equation for y we get v + t v￿^ = F (v) ⇒ v￿^ = ° F (v) − v ¢ t . This last equation is separable. This establishes the Theorem. §

Example 1.3.9: Determine whether the equation below is homogeneous, y ￿ = t^2 1 − y^3

Solution: Dividing numerator and denominator by t^3 we obtain y￿^ = t^2 (1 − y^3 )

t^3

t^3 ¥ ⇒^ y

t

t^3

y t

We conclude that the differential equation is not homogeneous. Theorem 1.3.5 (Homogeneous equations). If the differential equation y￿(t) is homogeneous, then the differential equation for the unknown v(t) = y(t) t is s The Theorem says that homogeneous equations can be transformed into separab If the differential equation for y is homogeneous, then the differential equation is separable. Therefore, an homogeneous equation for y is solved as follows: F separable equation for v; second, solve for v in implicit or explicit form; third, y = t v. Proof of Theorem 1.3.5: If y￿^ = f (t, y) is homogeneous, then it can b y￿^ = F (y/t) for some function F. Introduce v = y/t. This means, y(t) = t v(t) ⇒ y￿(t) = v(t) + t v￿(t). Introducing these expressions into the differential equation for y we get v + t v￿^ = F (v) ⇒ v￿^ =

F (v) − v

t

This last equation is separable. This establishes the Theorem. 2 2 G. NAGY – ODE January 26, 2012 21

C

1.3.9: Determine whether the equation below is homogeneous, y￿^ = t^2 1 − y^3

n: Dividing numerator and denominator by t^3 we obtain y￿^ = t^2 (1 − y^3 )

t^3

t^3 ¥ ⇒^ y

t

t^3

y t

de that the di fferential equation is not homogeneous. C 1.3.5 (Homogeneous equations). If the differential equation y￿(t) = f

t, y(t)

neous, then the di fferential equation for the unknown v(t) = y(t) t is separable. rem says that homogeneous equations can be transformed into separable equations. erential equation for y is homogeneous, then the differential equation for v = y/t le. Therefore, an homogeneous equation for y is solved as follows: First, find the equation for v; second, solve for v in implicit or explicit form; third, replace back heorem 1.3.5: If y￿^ = f (t, y) is homogeneous, then it can be written as t) for some function F. Introduce v = y/t. This means, y(t) = t v(t) ⇒ y￿(t) = v(t) + t v￿(t). ng these expressions into the differential equation for y we get ￿ ￿

F (v) − v

C

Example 1.3.9: Determine whether the equation below is homogeneous, y

￿

=

t^2 1 − y^3 . Solution: Dividing numerator and denominator by t^3 we obtain y

￿

=

t^2 (1 − y^3 ) ≥ 1 t^3 ¥ ≥ 1 t^3 ¥ ⇒^ y

￿

=

≥ 1 t ¥ ≥ 1 t^3 ¥ − ≥ y t ¥ 3. We conclude that the differential equation is not homogeneous. C Theorem 1.3.5 (Homogeneous equations). If the differential equation y￿(t) = f ° t, y(t) ¢ is homogeneous, then the differential equation for the unknown v(t) = y(t) t is separable. The Theorem says that homogeneous equations can be transformed into separable equations. If the differential equation for y is homogeneous, then the differential equation for v = y/t is separable. Therefore, an homogeneous equation for y is solved as follows: First, find the separable equation for v; second, solve for v in implicit or explicit form; third, replace back y = t v. Proof of Theorem 1.3.5: If y￿^ = f (t, y) is homogeneous, then it can be written as y￿^ = F (y/t) for some function F. Introduce v = y/t. This means, y(t) = t v(t) ⇒ y

￿

(t) = v(t) + t v

￿

(t).

Introducing these expressions into the differential equation for y we get v + t v

￿

= F (v) ⇒ v

￿

=

° F (v) − v ¢ t .

Example 1.3.9: Determine whether the equation below is homogeneous, y￿^ = t^2 1 − y^3

Solution: Dividing numerator and denominator by t^3 we obtain y￿^ = t^2 (1 − y^3 )

t^3

t^3 ¥ ⇒^ y

t

t^3

y t

We conclude that the differential equation is not homogeneous. Theorem 1.3.5 (Homogeneous equations). If the differential equation y￿( is homogeneous, then the differential equation for the unknown v(t) = y(t) t is The Theorem says that homogeneous equations can be transformed into separa If the differential equation for y is homogeneous, then the differential equatio is separable. Therefore, an homogeneous equation for y is solved as follows: separable equation for v; second, solve for v in implicit or explicit form; third y = t v. Proof of Theorem 1.3.5: If y￿^ = f (t, y) is homogeneous, then it can y￿^ = F (y/t) for some function F. Introduce v = y/t. This means, y(t) = t v(t) ⇒ y￿(t) = v(t) + t v￿(t). Introducing these expressions into the differential equation for y we get v + t v￿^ = F (v) ⇒ v￿^ =

F (v) − v

t

This last equation is separable. This establishes the Theorem. Example 1.3.10: Find all solutions y of the ODE y￿^ = t^2 + 3y^2 2 ty

For a homogeneous equation ↔ substitute ⟹ Separable equation

v + t v￿^ = F (v) ⇒ v￿^ =

F (v) − v

t

This last equation is separable. This establishes the Theorem. §

Example 1.3.10: Find all solutions y of the ODE y

￿

t^2 + 3y^2

2 ty

Solution: The equation is homogeneous, since

y

￿

(t^2 + 3y^2 )

2 ty

t^2

t^2

¥ ⇒^ y

￿

y

t

y

t

Therefore, we introduce the change of unknown v = y/t, so y = t v and y￿^ = v + t v￿. Hence

v + t v

￿

1 + 3v^2

2 v

⇒ t v

￿

1 + 3v^2

2 v

− v =

1 + 3v^2 − 2 v^2

2 v

We obtain the separable equation v

￿

t

1 + v^2

2 v

. We rewrite and integrate it,

2 v

1 + v^2

v￿^ =

t

Z

2 v

1 + v^2

v￿^ dt =

Z

t

dt + c 0.

The substitution u = 1 + v^2 (t) implies du = 2v(t) v￿(t) dt, so Z du

u

Z

dt

t

  • c 0 ⇒ ln(u) = ln(t) + c 0 ⇒ u = eln(t)+c^0.

v + t v = F (v) ⇒ v =

t

This last equation is separable. This establishes the Theorem. §

Example 1.3.10: Find all solutions y of the ODE y

￿

t^2 + 3y^2

2 ty

Solution: The equation is homogeneous, since

y

￿

(t^2 + 3y^2 )

2 ty

t^2

t^2

¥ ⇒^ y

￿

y

t

y

t

Therefore, we introduce the change of unknown v = y/t, so y = t v and y￿^ = v + t v￿. Hence

v + t v￿^ =

1 + 3v^2

2 v

⇒ t v￿^ =

1 + 3v^2

2 v

− v =

1 + 3v^2 − 2 v^2

2 v

We obtain the separable equation v

￿

t

1 + v^2

2 v

. We rewrite and integrate it,

2 v

1 + v^2

v￿^ =

t

Z

2 v

1 + v^2

v￿^ dt =

Z

t

dt + c 0.

The substitution u = 1 + v^2 (t) implies du = 2v(t) v￿(t) dt, so

Z

du

u

Z

dt

t

+ c 0 ⇒ ln(u) = ln(t) + c 0 ⇒ u = e

ln(t)+c 0

Example: