Solve DE: $y'' + y' = x$ by letting $p = y'$.

Question

Solve DE: $y'' + y' = x$ by letting $p = y'$.

Doesn't seem to work for exact equations or homogenous solutions.

Answer 1

With factor $e^x$: $$(e^xy')'=e^xy''+e^xy'=e^x(y''+y')=xe^x$$ then with integration $$e^xy'=e^x(x-1)+C_1$$ or $$y'=x-1+C_1e^{-x}$$ and finally $$y=\dfrac12x^2-x-C_1e^{-x}+C_2$$