Solving the IVP $y'+2xy=x^2,\,y(0)=3$.

Question

Solve the following IVP and leave the answer in a form involving a definite integral. $$\left\{\begin{align*}y'+2xy&=x^2\\y(0)&=3\end{align*}\right.$$

I have tried integrating factor but ı got stuck at $(ye^{x^2})'=(x^2)\cdot e^{x^2}$

Answer 1

This is a linear equation... Multiplying by the integrating factor $e^{\int_0^x 2t dt}$ will get you the solution: \begin{align*} y(x)= &e^{-\int_0^x 2t dt}\left(y(0) + \int_0^x \left(t^2 e^{\int_0^t 2s ds} \right) dt \right)=e^{-x^2} \left(3 + \int_0^x t^2 e^{t^2}dt \right)\\ = & 3 e^{-x^2} + e^{-x^2} \int_0^x t^2 e^{t^2} dt \end{align*}

However, we end up with an indefinite integral, not a definite one.

Answer 2

Starting from your last line: $$(ye^{x^2})'=(x^2)\cdot e^{x^2}$$ $$\int_?^? d(ye^{x^2})=\int_0^xt^2 e^{t^2}dt$$ $$\int_?^v dv=\int_0^xt^2 e^{t^2}dt$$ For the bounds of the integral on the LHS we have: $$v=ye^{x^2}=y(0)e^{0^2}=3$$ $$v-3=\int_0^xt^2 e^{t^2}dt$$ $$ye^{x^2}-3=\int_0^xt^2 e^{t^2}dt$$ Finally: $$y(x)=3e^{-x^2}+e^{-x^2}\int_0^x t^2 e^{t^2}dt$$