# Help: Solution to $y' + 2x^2y = x^2$

Hi, I have been going through the courses for differential equations and just got done with the section on First Order DE. I decided to practice a few more problems on my own and I have been stumped for a while now because when I tried solving it I get different results and I cannot figure out what I am doing wrong. I think it's going to end up being such a small detail I missed but I cannot see it.

I was trying to solve it the way I saw here on Brilliant for a nonhomogeneous equation using the formula $y(x) = A(x)e^{-\int p(x)dx}$ where $A(x) = \int q(x)e^{\int p(x)dx}$ . When I solve for $A(x)$ I get $A(x) = \int x^2e^{\frac{2}{3}x^3}dx$ then after u-substitution I get $A(x) = \frac{1}{2}e^{\frac{2}{3}x^3}$. Going back to the other formula $y(x) = A(x)e^{-\int p(x)dx}$ then means $y(x) = \frac{1}{2}e^{\frac{2}{3}x^3}e^{-\frac{2}{3}x^3}$ which is $\frac{1}{2}$ which is not correct. The way the book does it is by seperation of variables but what am I doing wrong with using the above formulas?

Note by Christian Bracamontes
2 years, 3 months ago

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