Show that the differential equation of which $y = 2 (x^{2} - 1) + c e^{- x^{2}}$ is a solution is $\frac{d y}{d x} + 2 x y = 4 x^{3}$

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Solution

We have,
$y = 2 (x^{2} - 1) + c e^{- x^{2}}$ ...(1)
Differentiating both sides of (1) with respect to $x$ , we get
$\frac{d y}{d x} = 4 x - c e^{- x^{2}} 2 x = 2 x [2 - c e^{- x^{2}}] = - 2 x [2 x^{2} - 2 + c e^{- x^{2}} - 2 x^{2}] = - 2 x [2 (x^{2} - 1) + c e^{- x^{2}} - 2 x^{2}] = - 2 x [y - 2 x^{2}] [Using (1)] \Rightarrow \frac{d y}{d x} = - 2 x y + 4 x^{3} \Rightarrow \frac{d y}{d x} + 2 x y = 4 x^{3}$
Hence, the given function is the solution to the given differential equation.

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