Integrating factor mistakes when solving 1 order ODE

Question

I have an ODE:
$$\frac{dy}{dx} + 3x^{2}y = x^{2}$$ .

I got the following integrating factor:

$$e^{x^3}$$

Then I multiplied both sides, but didn't come up with the right answer. It should be:

$$y = c~e^{-x^3} + \frac{1}{3}. $$

When I come up with the following:

$$y = \left(\frac{1}{3}e^{x^3} + c\right) e^{x^3}$$

What am I doing wrong?

Answer 1

You have: $$ y'(x)+3x^2 y-x^2=0\tag{1} $$ The associated homogeneous equation $$ y'=-3x^2 y $$ gives you: $$ \frac{y'}{y}=-3x^2=\frac{\mathrm{d}}{\mathrm{d}x}\log(y) $$ so $$ \log (y)=-x^3, $$ and $$ y(x)=e^{-x^3}. $$ The general integral for the homogeneous equation is $$ \left\{c\,e^{-x^3},\ c\in\mathbb{R}\right\}. $$ Now I look for a particular solution $\bar{y}$ of the original equation. If I look for a constant, like $$ \bar{y}(x)=a, $$ it must be $\bar{y}'(x)=0$, hence, from $(1)$: $$ 0+3ax^2-x^2=0. $$ It follows $a=\frac{1}{3}$ and the solution: $$ y(x)=ce^{-x^3}+\frac{1}{3}, $$ for all $c\in\mathbb{R}$.

Answer 2

Your LHS should read $$\frac{d}{dx}(e^{x^3}y),$$ so after integrating both sides, you have to divide by $e^{x^3}$ (i.e. multiply by $e^{-x^3}$) to obtain $y(x)$.

Answer 3

You found the right integrating factor, allowing us to write the ODE as $$(e^{x^3} y)'=e^{x^3}x^2,$$ by multiplying both sides by $e^{x^3}.$ Integrating both sides gives $$ e^{x^3} y=\frac{e^{x^3}}{3}+c.$$ Now, just multiply both sides by $e^{-x^3},$ which gives $$y(x)=\frac{1}{3}+ce^{-x^3},$$ since $e^{-x^3}e^{x^3}=1.$ It looks like you should've multiplied both sides by $e^{-x^3}$ but did $e^{x^3}$, by mistake.

Answer 4

You have probably forgotten to put minus sign when dividing by $e^{x^3}$.

$$e^{x^3}dy + 3x^2e^{x^3}dx = e^{x^3}x^2dx\\ d(e^{x^3}y)=\frac13d(e^{x^3})\\ e^{x^3}y=\frac13e^{x^3}+C\\ y=Ce^{-x^3}+\frac13$$