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 $.
$ 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 $.
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