Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 13 / b

For $x\ne 0$, multiply by $x$ to obtain the first-order linear ordinary differential equation $y^{\prime }+xy=5x{e}^{-x^2}$. Its integrating factor is $\mu =e^{x^2/2}$. The product rule then givesIntegration yields $y{e}^{x^2/2}=-5e^{-x^2/2}+C$, soThe constant of integration is arbitrary on each connected interval of the original equation's domain. The formula extends smoothly through zero, but the printed equation itself contains $1/x$ and is not defined there.

The real equation has domain $x>0$. Its integrating factor isThus the linear ordinary differential equation becomes $(x^{x}y{)}^{\prime }=x^x{x}^{-x}=1$. After integration, $x^{x}y=x+C$. The initial condition gives $2=1+C$, hence