Verifying differential equation solutions.

Verify that \[ y = \frac{1}{2} + ce^{-x^2} \] is a solution to \[ y' + 2xy = x \qquad (1) \]

Calculate y'.

\[ \begin{align} y' &= \frac{d}{dx}(\frac{1}{2} + ce^{-x^2}) \\ &= -2xce^{-x^2} \end{align} \ \]

Substitute y' into equation (1).

\[ \begin{align} y' + 2xy &= -2xce^{-x^2} + 2x(\frac{1}{2} + ce^{-x^2}) \\ &= -2xce^{-x^2} + x + 2xce^{-x^2} \\ &= x \end{align} \ \] so the given y is a solution.

You will gain most benefit if you work through each problem first and then come back and check your work against the solution.

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Now You Try:

Verify that \[ y = \cos x + \sin x \] is a solution to \[ y' + y = 2 \cos x \]