540 CHAPTER 29. FIRST ORDER SCALAR ODE

2. Multiply by exp(A(t))

chain rule and product rule︷ ︸︸ ︷exp(A(t))

(y′+a(t)y

)=

ddt

(exp(A(t))y) = exp(A(t))b(t)

3. Do∫

to both sides. Pick F (t) ∈∫

exp(A(t))b(t)dt.

exp(A(t))y(t) = F (t)+C

y(t) = exp(−A(t))F (t)+C exp(−A(t))

This proves the following theorem.

Theorem 29.1.7 The solutions to the equation, y′+ a(t)y = b(t) consist of all functionsof the form y(t) = e−A(t)F (t)+ e−A(t)C where F (t) ∈

∫eA(t)b(t)dt and C is a constant,

A′ (t) = a(t).

Finally, here is a uniqueness theorem.

Theorem 29.1.8 If a(t) is a continuous function, there is at most one solution to the initialvalue problem, y′+a(t)y = b(t) , y(r) = y0.

Proof: If there were two solutions y1 and y2, then letting w = y1− y2, it follows w′+a(t)w = 0 and w(r) = 0. Then multiplying both sides of the differential equation by eA(t)

where A′ (t) = a(t) , it follows(

eA(t)w)′

= 0 and so eA(t)w(t) = C for some constant, C.

However, w(r) = 0 and so this constant can only be 0. Hence w = 0 and so y1 = y2. ■Finally, consider the general linear initial value problem.

Definition 29.1.9 A linear differential equation is one which is of the form

y′+a(t)y = b(t)

where a,b are continuous. The corresponding initial value problem is

y′+a(t)y = b(t) , y(t0) = y0.

Now here are the steps for solving the initial value problem.

1. Find the integrating factor∫

a(t)dt ≡A(t)+C. The integrating factor is exp(A(t))=eA(t).

2. Multiply both sides by the integrating factor.

exp(A(t))(y′ (t)+a(t)y(t)

)=

ddt

(exp(A(t))y(t)) = exp(A(t))b(t)

Why is this so? It involves the chain rule and the product rule.

ddt

(exp(A(t))y(t)) = exp(A(t))A′ (t)y(t)+ exp(A(t))y′ (t)

= exp(A(t))a(t)y(t)+ exp(A(t))y′ (t)

= exp(A(t))(y′ (t)+a(t)y(t)

)