The problem is: $$\frac{dy}{dx} +{3x^2}{y} = 6x^2.$$

The part I am confused about is after you get your integrating factor which is:

$$I(x) = e^{\int 3x^2} \, dx = e^{{x}^{3}}$$

and multiplying both sides of the differential equation by $e^{{x}^{3}}$, you get:

  1. $$e^{{x}^{3}}\frac{dy}{dx} +{3x^2}{y}=6x^2e^{{x}^{3}}$$


  1. $$\frac{d}{dx}(e^{{x}^{3}}y)$$

I don't understand how to get from 1) to 2). If someone could please explain this to me, I would appreciate it. Thank you!

  • $\begingroup$ (1) is not correct, you need to be more careful with your algebra. Once you've fixed it, don't try to get from (1) to (2), get from (2) to (1), it's easier. $\endgroup$
    – David
    Jul 30, 2023 at 5:56

3 Answers 3


Actually the integratic factor is not $(e^x)^3$ because $(e^x)^3=e^{3x}$

It is $I(x)=e^{x^3}$

Then for step 1) :


Multiplying the whole differential equation by $I(x)$.

$$ e^{x^3}\dfrac{dy}{dx}+3x^2e^{x^3}y=6x^2e^{x^3} \ (1) $$

Write then the general form $\dfrac{d}{dx}(fy)=\dfrac{df}{dx}y+f\dfrac{dy}{dx}$and identify this form in the LHS of the equation $(1)$

Because $\dfrac{d(e^{x^3})}{dx}=3x^2e^{x^3}$

It gives $\dfrac{d(e^{x^3}y)}{dx}$

So going to the 2)


$$\dfrac{d(e^{x^3}y)}{dx}=2 \times 3x^2e^{x^3} =2\dfrac{d(e^{x^3})}{dx}$$

Now you can integrate over for example $[0,x]$ for a given $x \in \mathbb{R}$

$$ e^{x^3}y(x)-1\cdot y(0)=2e^{x^3}-2\cdot1$$

So after precising your initial conditions $b=y(0)$

$$ y : x \to 2 + (b-2)e^{-x^3} $$

You can verify by hand that it well satisfy the initial equation independently ofr $b$.

  • $\begingroup$ I updated the question to fit your comment. Can you please go into detail now that it's fixed going from 1) to 2)? Thank you very much $\endgroup$ May 19, 2023 at 20:23
  • 1
    $\begingroup$ @JacobAvenaim It is good now for you ? $\endgroup$
    – EDX
    May 19, 2023 at 20:38

By the looks of the differential equation, the integrating factor is $e^{x^3}$ and not $(e^{x})^{3}=e^{3x}$.

Now if you correct that, notice that $(e^{x^3}y)'=e^{x^3}y'+3x^2e^{x^3}y$.

So you go from:

  1. $e^{x^{3}}\frac{dy}{dx} +{3x^2}e^{x^{3}}{y}=6x^2e^{x^{3}}$


  1. $\frac{d}{dx}(e^{x^{3}}y)=6x^2e^{x^{3}}$

The DE is separable no need for an integrating factor: $$\frac{dy}{dx} +{3x^2}{y} = 6x^2$$ $$\frac{dy}{dx} =3x^2(2-y)$$ $$\int \dfrac {dy}{y-2}=-3\int x^2dx$$


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