Verifying differential equation solutions.

Verify that \[ y = \frac{1}{1+x^2} \] is a solution to \[ y' + 2xy^2 = 0 \qquad (1) \]

Calculate y'.

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

Substitute y' into equation (1).

\[ \begin{align} y' + 2xy^2 &= - \frac{2x}{(1+x^2)^2} + 2x\frac{1}{(1+x^2)^2} \quad \\ &= 0 \quad \\ \end{align} \ \] so the given y is a solution.

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Now You Try:

Verify that \[ y = e^{-2x} \] is a solution to \[ y'' + 4y' + 4y = 0 \]