8 Solving equations via the integrating factor

Having found the integrating factor for a linear equation we now proceed to solve the equation.

Returning to the differential equation, written in standard form:

d i d t + R L i = E L cos ω t

for which the integrating factor is

e R t L

we multiply the equation by the integrating factor to obtain

e R t L d i d t + R L e R t L i = E L e R t L cos ω t

At this stage the left-hand side of this equation can always be simplified as follows:

d d t ( e R t L i ) = E L e R t L cos ω t .

Now this is in the form of an exact differential equation and so we can integrate both sides to obtain the solution:

e R t L i = E L e R t L cos ω t d t .

All that remains is to complete the integral on the right-hand side. Using the method of integration by parts we find

e R t L cos ω t d t = L L 2 ω 2 + R 2 ω L sin ω t + R cos ω t e R t L

Hence

e R t L i = E L 2 ω 2 + R 2 ω L sin ω t + R cos ω t e R t L + C .

Finally

i = E L 2 ω 2 + R 2 ω L sin ω t + R cos ω t + C e R t L .

is the solution to the original differential equation (1). Note that, as we should expect for the solution to a first order differential equation, it contains a single arbitrary constant C .

Task!

Using the integrating factors found earlier in the Task on pages 22-23, find the general solutions to the differential equations

  1. x 2 d y d x + 2 x 2 y = x 2 e 2 x
  2. t 2 d i d t + 2 t 2 i = t 2 e 2 t
  3. d y d x ( tan x ) y = 1.
  1. The standard form is d y d x + 2 y = e 2 x for which the integrating factor is e 2 x . e 2 x d y d x + 2 e 2 x y = 1 i.e. d d x ( e 2 x y ) = 1 so that  e 2 x y = x + C leading to y = ( x + C ) e 2 x
  2. The general solution is i = ( t + C ) e 2 t as this problem is the same as (1) with different variables.
  3. The equation is in standard form and the integrating factor is cos x . then d d x ( cos x y ) = cos x so that  cos x y = cos x d x = sin x + C giving y = tan x + C sec x