Reduction of order (I)

Problem:

$y'' - y' = x$

Solution:

y is missing from the equation, so we can simply let $z = y'$ and so $y'' = z'$. The equation is reduced to

$\frac{dz}{dx} - z = x$

This is a first order linear equation, by inspection, the integrating factor is $e^{-x}$, we have

$\frac{d}{dx}(e^{-x} z) = e^{-x} \frac{dz}{dx} - e^{-x} z = e^{-x} (\frac{dz}{dx} - z) = x e^{-x}$.

Therefore $e^{-x} z = \int {x e^{-x} dx } = - xe^{-x} - e^{-x} + C_1$.

Then we have $z = -x - 1 + C_1e^x$

Last but not least, $y = \int z dx = -\frac{x^2}{2} - x + C_1e^x + C_2$.