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Monday, December 28, 2015

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 $.

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