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This is an introduction to ordinary differential equations. We describe the main ideas to solve certain differential equations, ...
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Mathematics Department, Michigan State University, East Lansing, MI, 48824. gnagy@msu.edu
January 18, 2021
x 2
x 1
x^1
x^2
b^ a
0
First Order Equations
We start our study of differential equations in the same way the pioneers in this field did. We show particular techniques to solve particular types of first order differential equations. The techniques were developed in the eighteenth and nineteenth centuries and the equations include linear equations, separable equations, Euler homogeneous equations, and exact equa- tions. This way of studying differential equations reached a dead end pretty soon. Most of the differential equations cannot be solved by any of the techniques presented in the first sections of this chapter. People then tried something different. Instead of solving the equa- tions they tried to show whether an equation has solutions or not, and what properties such solution may have. This is less information than obtaining the solution, but it is still valu- able information. The results of these efforts are shown in the last sections of this chapter. We present theorems describing the existence and uniqueness of solutions to a wide class of first order differential equations.
t
y
π 2
π 2
y′^ = 2 cos(t) cos(y)
3
1.1. LINEAR CONSTANT COEFFICIENT EQUATIONS 5
space variables, and the heat equation in example (c) is first order in time and second order in space variables.
1.1.2. Linear Differential Equations. We start with a precise definition of a first order ordinary differential equation. Then we introduce a particular type of first order equations—linear equations.
Definition 1.1.1. A first order ODE on the unknown y is
y′(t) = f (t, y(t)), (1.1.1)
where f is given and y′^ = dy dt
. The equation is linear iff the source function f is linear on its second argument, y′^ = a(t) y + b(t). (1.1.2) The linear equation has constant coefficients iff both a and b above are constants. Oth- erwise the equation has variable coefficients.
There are different sign conventions for Eq. (1.1.2) in the literature. For example, Boyce- DiPrima [ 3 ] writes it as y′^ = −a y + b. The sign choice in front of function a is matter of taste. Some people like the negative sign, because later on, when they write the equation as y′^ + a y = b, they get a plus sign on the left-hand side. In any case, we stick here to the convention y′^ = ay + b.
Example 1.1.2. (a) An example of a first order linear ODE is the equation
y′^ = 2 y + 3.
On the right-hand side we have the function f (t, y) = 2y + 3, where we can see that a(t) = 2 and b(t) = 3. Since these coefficients do not depend on t, this is a constant coefficient equation. (b) Another example of a first order linear ODE is the equation
y′^ = −
t
y + 4t.
In this case, the right-hand side is given by the function f (t, y) = − 2 y/t + 4t, where a(t) = − 2 /t and b(t) = 4t. Since the coefficients are nonconstant functions of t, this is a variable coefficients equation. (c) The equation y′^ = −
ty
C
We denote by y : D ⊂ R → R a real-valued function y defined on a domain D. Such a function is solution of the differential equation (1.1.1) iff the equation is satisfied for all values of the independent variable t on the domain D.
Example 1.1.3. Show that y(t) = e^2 t^ −
is solution of the equation
y′^ = 2 y + 3.
6 1. FIRST ORDER EQUATIONS
Solution: We need to compute the left and right-hand sides of the equation and verify they agree. On the one hand we compute y′(t) = 2e^2 t. On the other hand we compute
2 y(t) + 3 = 2
e^2 t^ −
We conclude that y′(t) = 2 y(t) + 3 for all t ∈ R. C
Example 1.1.4. Find the differential equation y′^ = f (y) satisfied by y(t) = 4 e^2 t^ + 3. Solution: (Solution Video) We compute the derivative of y, y′^ = 8 e^2 t We now write the right-hand side above, in terms of the original function y, that is, y = 4 e^2 t^ + 3 ⇒ y − 3 = 4 e^2 t^ ⇒ 2(y − 3) = 8 e^2 t. So we got a differential equation satisfied by y, namely y′^ = 2y − 6. C
1.1.3. Solving Linear Differential Equations. Linear equations with constant co- efficient are simpler to solve than variable coefficient ones. But integrating each side of the equation does not work. For example, take the equation y′^ = 2 y + 3, and integrate with respect to t on both sides, ∫ y′(t) dt = 2
y(t) dt + 3t + c, c ∈ R.
The Fundamental Theorem of Calculus implies y(t) =
y′(t) dt, so we get
y(t) = 2
y(t) dt + 3t + c.
Integrating both sides of the differential equation is not enough to find a solution y. We still need to find a primitive of y. We have only rewritten the original differential equation as an integral equation. Simply integrating both sides of a linear equation does not solve the equation. We now state a precise formula for the solutions of constant coefficient linear equations. The proof relies on a new idea—a clever use of the chain rule for derivatives.
Theorem 1.1.2 (Constant Coefficients). The linear differential equation y′^ = a y + b (1.1.3) with a 6 = 0, b constants, has infinitely many solutions,
y(t) = c eat^ − b a
, c ∈ R. (1.1.4)
Remarks: (a) Equation (1.1.4) is called the general solution of the differential equation in (1.1.3).
(b) Theorem 1.1.2 says that Eq. (1.1.3) has infinitely many solutions, one solution for each value of the constant c, which is not determined by the equation.
8 1. FIRST ORDER EQUATIONS
We now compute exponentials on both sides, to get
y˜(t) = ±e^2 t+c^0 = ±e^2 t^ ec^0 , denote c = ±ec^0 , then y˜(t) = c e^2 t, c ∈ R.
Since ˜y = y +
, we get y(t) = c e^2 t^ −
, where c ∈ R. C
Remark: We converted the original differential equation y′^ = 2 y + 3 into a total derivative of a potential function ψ′^ = 0. The potential function can be computed from the step
ln(|y˜|)′^ = 2 ⇒
ln(|y˜|) − 2 t
then a potential function is ψ(t, y(t)) = ln
∣y(t) +^3 2
− 2 t. Since the equation is now
ψ′^ = 0, all solutions are ψ = c 0 , with c 0 ∈ R. That is
ln
∣y(t) +^3 2
− 2 t = c 0 ⇒ ln
∣y(t) +^3 2
= 2t + c 0 ⇒ y(t) +
= ±e^2 t+c^0.
If we denote c = ±ec^0 , then we get the solution we found above, y(t) = c e^2 t^ −
1.1.4. The Integrating Factor Method. The argument we used to prove Theo- rem 1.1.2 cannot be generalized in a simple way to all linear equations with variable coef- ficients. However, there is a way to solve linear equations with both constant and variable coefficients—the integrating factor method. Now we give a second proof of Theorem 1.1. using this method.
Second Proof of Theorem 1.1.2: Write the equation with y on one side only,
y′^ − a y = b,
and then multiply the differential equation by a function μ, called an integrating factor,
μ y′^ − a μ y = μ b. (1.1.5)
Now comes the critical step. We choose a positive function μ such that
− a μ = μ′. (1.1.6)
For any function μ solution of Eq. (1.1.6), the differential equation in (1.1.5) has the form
μ y′^ + μ′^ y = μ b.
But the left-hand side is a total derivative of a product of two functions, ( μ y
= μ b. (1.1.7)
This is the property we want in an integrating factor, μ. We want to find a function μ such that the left-hand side of the differential equation for y can be written as a total derivative, just as in Eq. (1.1.7). We only need to find one of such functions μ. So we go back to Eq. (1.1.6), the differential equation for μ, which is simple to solve,
μ′^ = −a μ ⇒
μ′ μ
= −a ⇒
ln(|μ|)
= −a ⇒ ln(|μ|) = −at + c 0.
Computing the exponential of both sides in the equation above we get
μ = ±ec^0 −at^ = ±ec^0 e−at^ ⇒ μ = c 1 e−at, c 1 = ±ec^0.
Since c 1 is a constant which will cancel out from Eq. (1.1.5) anyway, we choose the integration constant c 0 = 0, hence c 1 = 1. The integrating function is then
μ(t) = e−at.
1.1. LINEAR CONSTANT COEFFICIENT EQUATIONS 9
This function is an integrating factor, because if we start again at Eq. (1.1.5), we get
e−at^ y′^ − a e−at^ y = b e−at^ ⇒ e−at^ y′^ +
e−at
y = b e−at,
where we used the main property of the integrating factor, −a e−at^ =
e−at
. Now the product rule for derivatives implies that the left-hand side above is a total derivative, ( e−at^ y
= b e−at.
The right-hand side above can be rewritten as a derivative, b e−at^ =
b a e−at
, hence ( e−at^ y +
b a
e−at
y +
b a
e−at
We have succeeded in writing the whole differential equation as a total derivative. The differential equation is the total derivative of a potential function, which in this case is
ψ(t, y) =
y + b a
e−at.
Notice that this potential function is the exponential of the potential function found in the first proof of this Theorem. The differential equation for y is a total derivative,
dψ dt
(t, y(t)) = 0,
so it is simple to integrate,
ψ(t, y(t)) = c ⇒
y(t) +
b a
e−at^ = c ⇒ y(t) = c eat^ −
b a
This establishes the Theorem. We solve the example below following the second proof of Theorem 1.1.2.
Example 1.1.6. Find all solutions to the constant coefficient equation
y′^ = 2y + 3 (1.1.8)
Solution: (Solution Video) Write the equation in (1.1.8) as follows,
y′^ − 2 y = 3.
Multiply this equation by the integrating factor μ(t) = e−^2 t,
e−^2 ty′^ − 2 e−^2 t^ y = 3 e−^2 t^ ⇔ e−^2 ty′^ +
e−^2 t
y = 3 e−^2 t.
We now solve the same problem above, but now using the formulas in Theorem 1.1.2.
Example 1.1.7. Find all solutions to the constant coefficient equation
y′^ = 2y + 3 (1.1.9)
Solution: The equation above is the case of a = 2 and b = 3 in Eq. (1.1.3). Therefore, using these values in the expression for the solution given in Eq. (1.1.4) we obtain
y(t) = c e^2 t^ −
1.1. LINEAR CONSTANT COEFFICIENT EQUATIONS 11
Remark: In case t 0 = 0 the initial condition is y(0) = y 0 and the solution is
y(t) =
y 0 +
b a
eat^ −
b a
The proof of Theorem 1.1.4 is just to write the general solution of the differential equation given in Theorem 1.1.2, and fix the integration constant c with the initial condition.
Proof of Theorem 1.1.4: The general solution of the differential equation in (1.1.10) is given in Eq. (1.1.4) for any choice of the integration constant c,
y(t) = c eat^ −
b a
The initial condition determines the value of the constant c, as follows
y 0 = y(t 0 ) = c eat^0 −
b a
⇔ c =
y 0 +
b a
e−at^0.
Introduce this expression for the constant c into the differential equation in Eq. (1.1.10),
y(t) =
y 0 + b a
ea(t−t^0 )^ − b a
This establishes the Theorem.
Example 1.1.8. Find the unique solution of the initial value problem
y′^ = 2y + 3, y(0) = 1. (1.1.13)
Solution: (Solution Video) All solutions of the differential equation are given by
y(t) = ce^2 t^ −
where c is an arbitrary constant. The initial condition in Eq. (1.1.13) determines c,
1 = y(0) = c −
⇒ c =
Then, the unique solution to the initial value problem above is y(t) =
e^2 t^ −
Example 1.1.9. Find the solution y to the initial value problem
y′^ = − 3 y + 1, y(0) = 1.
Solution: (Solution Video) Write the differential equation as y′^ + 3 y = 1, and multiply the equation by the integrating factor μ = e^3 t, which will convert the left-hand side above into a total derivative,
e^3 ty′^ + 3 e^3 t^ y = e^3 t^ ⇔ e^3 ty′^ +
e^3 t
y = e^3 t.
This is the key idea, because the derivative of a product implies ( e^3 t^ y
= e^3 t.
The exponential e^3 t^ is an integrating factor. Integrate on both sides of the equation,
e^3 t^ y =
e^3 t^ + c.
So every solution of the differential equation above is given by
y(t) = c e−^3 t^ +
, c ∈ R.
12 1. FIRST ORDER EQUATIONS
The initial condition y(0) = 2 selects only one solution,
1 = y(0) = c +
⇒ c =
We get the solution y(t) =
e−^3 t^ +
Notes. This section corresponds to Boyce-DiPrima [ 3 ] Section 2.1, where both constant and variable coefficient equations are studied. Zill and Wright give a more concise exposition in [ 17 ] Section 2.3, and a one page description is given by Simmons in [ 10 ] in Section 2.10. The integrating factor method is shown in most of these books, but unlike them, here we emphasize that the integrating factor changes the linear differential equation into a total derivative, which is trivial to integrate. We also show here how to compute the potential functions for the linear differential equations. In § 1.4 we solve (nonlinear) exact equations and nonexact equations with integrating factors. We solve these equations by transforming them into a total derivative, just as we did in this section with the linear equations.