Find $\dfrac{d}{dx}e^{-x^{2}}$. Hence find $\displaystyle \int x^{3}e^{-x^{2}}\,dx$. By using the substitution $z = e^{x^{2}}y$, find the general solution of $\dfrac{dy}{dx}-2xy = x^{3}$, expressing $y$ in terms of $x$.

1. Find the derivative of $e^{-x^2}$ using the chain rule.

$\frac{d}{dx}\left(e^{-x^2}\right) = e^{-x^2} \cdot \frac{d}{dx}(-x^2) = -2xe^{-x^2}$

2. Set up the integral using integration by parts, isolating the component related to the derivative from the previous part.

$\int x^3 e^{-x^2}\,dx = \int x^2 \left(x e^{-x^2}\right)\,dx$
Using integration by parts, let $u = x^2$ and $\frac{dv}{dx} = x e^{-x^2}$.
Then $\frac{du}{dx} = 2x$ and $v = -\frac{1}{2}e^{-x^2}$ (deduced from Step 1).
$\int x^3 e^{-x^2}\,dx = x^2 \left(-\frac{1}{2}e^{-x^2}\right) - \int \left(-\frac{1}{2}e^{-x^2}\right)(2x)\,dx$

3. Evaluate the remaining integral to obtain the final answer.

$= -\frac{1}{2}x^2 e^{-x^2} + \int x e^{-x^2}\,dx$
Since $\int x e^{-x^2}\,dx = -\frac{1}{2}e^{-x^2}$ from the first part:
$= -\frac{1}{2}x^2 e^{-x^2} - \frac{1}{2}e^{-x^2} + C$
$= -\frac{1}{2}e^{-x^2}\left(x^2 + 1\right) + C$

1. Differentiate the substitution and link to the given DE

With $z = e^{-x^{2}}y$,
$\frac{dz}{dx} = e^{-x^{2}}\frac{dy}{dx} + y\cdot\frac{d}{dx}e^{-x^{2}} = e^{-x^{2}}\frac{dy}{dx} - 2xe^{-x^{2}}y = e^{-x^{2}}\left(\frac{dy}{dx} - 2xy\right).$
Since $\dfrac{dy}{dx} - 2xy = x^{3}$,
$\frac{dz}{dx} = e^{-x^{2}}\cdot x^{3} = x^{3}e^{-x^{2}}.$

2. Integrate to find z using part (i)

Integrating both sides and using the result of part (i),
$z = \int x^{3}e^{-x^{2}}\,dx = -\tfrac{1}{2}e^{-x^{2}}(x^{2}+1) + C.$

3. Back-substitute and make y the subject

Since $z = e^{-x^{2}}y$,
$e^{-x^{2}}y = -\tfrac{1}{2}e^{-x^{2}}(x^{2}+1) + C.$
Multiplying through by $e^{x^{2}}$,
$y = -\tfrac{1}{2}(x^{2}+1) + Ce^{x^{2}}.$