How do you use integrating factors to solve this differential equation:?

Question

xy' + (1-x)y = e^(2x) / x

and also y' + ytanx = 0

Answer 1

First, write the equation in standard form:

y' + (1-x)/x y = e^(2x)/x²

Now, the integrating factor is e^(∫(1-x)/x dx) = e^(∫1/x - 1 dx) = e^(ln x - x) = xe^(-x) (note that any antiderivative will work here, we don't need to bother with finding the most general one). So multiplying by this, we find:

xe^(-x) y' + (1-x)e^(-x) y = e^x/x

You will note that the expression on the left is nothing more than xe^(-x) dy/dx + d(xe^(-x))/dx y, and per the product rule, is thus d(xe^(-x)y)/dx. If you've found the integrating factor correctly, this will always happen. So integrating both sides yields:

xe^(-x)y = ∫e^x/x dx + C

And dividing by xe^(-x) yields the solution:

y = (∫e^x/x dx + C)e^x/x

(Note that we do not attempt to solve the integral, since e^x/x has no elementary antiderivative).

The second problem is done in the same way. The integrating factor will be e^(∫tan x dx) = e^(- ln (cos x)) = 1/cos x = sec x. So we have:

sec x y' + sec x tan x y = 0

Again, the expression on the left is simply an instance of the product rule, so integrating both sides yields:

sec x y = C

y = C cos x

And we are done.