Lecture Notes in Differential Equations - Bruce E. Shapiro

28 LESSON 4. LINEAR EQUATIONS
Exponentiating both sides of the equation
∫
∫
∫
pdt + C pdt
µ = e = e e C = C 1 e
pdt
(4.27)
where C 1 = e C and C is any constant. Since we want any function µ that
will work, we are free to choose our constant arbitrarily, e.g., pick C = 0
hence C 1 = 1, and we find that
µ(t) = e
∫
p(t)dt
(4.28)
has the properties we desire. We say the equation 4.28 is an integrating
factor for the differential equation 4.10. Since we have already chosen
the constant of integrate, we can safely ignore the constant of integration
when integrating p. To recap, the general solution of y ′ + py = q is given by
equation 4.18 whenever µ is given by 4.28. The particular solution of a given
initial value problem involving a linear ODE is then solved by substituting
the initial condition into the general solution obtained in this manner.
It is usually easier to memorize the procedure rather than the formula for
the solution (equation 4.18):
Method to Solve y ′ + p(t)y = q(t)
1. Computer µ = e ∫ p(t)dt and observe that µ ′ (t) = p(t)µ(t).
2. Multiply the ODE through by µ(t) giving
µ(t)y ′ + µ ′ (t)y = µ(t)q(t)
3. Observe that the left-hand side is precisely (d/dt)(µ(t)y).
4. Integrate both sides of the equation over t, remembering that
∫
(d/dt)(µy)dt = µ(t)y,
∫
µ(t)y =
q(t)ydt + C
5. Solve for y by dividing through by µ. Don’t forget the constant
on the right-hand side of the equation.
6. Use the initial condition to find the value of the constant, if this
is an initial value problem.