Question

The variables x and y satisfy the differential equation dy/dx= xe^x+y. It is given that y = 0 when x = 0.

Answer 1

Rewrite the ODE

$\dfrac{\mathrm dy}{\mathrm dx}=xe^x+y$

as

$\dfrac{\mathrm dy}{\mathrm dx}-y=xe^x$

Divide both sides by $e^x$:

$e^{-x}\dfrac{\mathrm dy}{\mathrm dx}-e^{-x}y=x$

The left side condenses to the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dx}\left[e^{-x}y\right]=x$

Integrate both sides and solve for $y$:

$e^{-x}y=\displaystyle\int x\,\mathrm dx=\frac12x^2+C$

$y=\dfrac12x^2e^x+Ce^x$

Given that $y=0$ when $x=0$, we find

$0=\dfrac120^2e^0+Ce^0\implies C=0$

so that the solution is

$\boxed{y(x)=\dfrac{x^2e^x}2}$

I'm not sure I understand the claim that $x$ must be less than 1...