Question

Find the solution of y'+2xy=x, with y(0)=-2.

Answer 1

$$y'+2xy=x$$We seek an integration factor \(\mu\) such that when multiplying we get \(\mu y'+2x\mu y=\mu y'+\mu' y=(\mu y)'\). Thus we seek a solution to \(\mu'=2x\mu\):$$\mu'=2x\mu\\\frac{d\mu}{dx}=2x\mu\\\frac1{\mu}d\mu=2xdx\\\int\frac1\mu d\mu=\int2x\,dx\\\log \mu=x^2\\\mu=e^{x^2}$$

Answer 2

Now, we multiply throughout by \(\mu=e^{x^2}\):$$e^{x^2}y'+2xe^{x^2}y=xe^{x^2}$$We observe (as we expected) that \(e^{x^2}y'+2xe^{x^2}y=(e^{x^2}y)'\) by the 'reverse' product rule:$$(e^{x^2}y)'=xe^{x^2}\\e^{x^2}y=\int xe^{x^2}\,dx$$

Answer 3

Observe we can evaluate that antiderivative pretty easily using a substitution; let \(u=x^2\) so \(\frac12du=x\,dx\):$$e^{x^2}y=\frac12\int e^u\,du=\frac12 e^{x^2}+C\\y=\frac12+Ce^{-x^2}$$

Answer 4

Now, let's impose our initial condition \(y(0)=-2\):$$y(0)=\frac12+Ce^{0}=\frac12+C\\-2=\frac12+C\\-\frac52=C$$thus our solution is given by \(y=\frac12-\frac52e^{-x^2}\)

Answer 5

Thank you so much.