Question

Show that the differential equation of which y = 2(x² − 1) + $c{e}^{-x^2}$ is a solution, is $\frac{dy}{dx}+2xy=4x^3$.

Answer 1

The given equation is
$y=2\left(x^2-1\right)+c{e}^{-x^2}$ ...(1)
where $c$ is a parameter.
As this equation has one arbitrary constant, we shall get a differential equation of first order.
Differentiating equation (1) with respect to $x$, we get
$$\begin{gathered} \frac{dy}{dx}=2\left(2x\right)+c{e}^{-x^2}(-2x) \\ \Rightarrow \frac{dy}{dx}=4x-2xc{e}^{-x^2}...\left(2\right) \end{gathered}$$
From (1) and (2), we get
$\frac{dy}{dx}=4x-2x\left[y-2x^2+2\right]$
$$\begin{gathered} \Rightarrow \frac{dy}{dx}=4x-2xy+4x^3-4x \\ \Rightarrow \frac{dy}{dx}+2xy=4x^3 \end{gathered}$$
Hence, $y=2\left(x^2-1\right)+c{e}^{-x^2}$ is the solution to the differential equation $\frac{dy}{dx}+2xy=4x^3$.