What is the solution to the differential equation $\frac {dy}{dx} + xy = x$, when $y(0)=-6$?

Question

I know about implicit differentiation and integrals, but how do I solve this type of equations?

Answer 1

Multiply the equation with $e^{x^2/2}$: $$e^{x^2/2} y' + x e^{x^2/2} y = x e^{x^2/2}.$$

Now the left hand side can be written as a derivative: $$e^{x^2/2} y' + x e^{x^2/2} y = (e^{x^2/2} y)'.$$

Thus we have, $$(e^{x^2/2} y)' = x e^{x^2/2}.$$

We can now take the antiderivative of this: $$e^{x^2/2} y = e^{x^2/2} + C,$$ where $C$ is some constant.

Then we multiply the equation with $e^{-x^2/2}$: $$y = 1 + C e^{-x^2/2}.$$

Finally we use the condition $y(0) = -6$: $$-6 = y(0) = 1 + C,$$ i.e. $$C = -7.$$

Thus we arrive at the solution, $$y(x) = 1 - 7 e^{-x^2/2}.$$

Answer 2

As said by Minus One-Twelfth any differential equation of the form $$y'+p(x)y=q(x)$$ can be solved by multiplying through by an 'integrating factor' given by $$IF=\exp{\left(\int p(x) dx\right)}$$ which gives the new equation $$\exp{\left(\int p(x) dx\right)}y'+p(x)\exp{\left(\int p(x) dx\right)}y=q(x)\exp{\left(\int p(x) dx\right)}$$ $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\exp{\left(\int p(x) dx\right)}y\right)=q(x)\exp{\left(\int p(x) dx\right)}$$ $$\therefore y=\exp{\left(-\int p(x) dx\right)}\int q(x)\exp{\left(\int p(x) dx\right)}dx$$