Solving $\frac{dy}{dx}+3x^2y=6x^2$ using integrating factor

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

or

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!

• (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. Jul 30, 2023 at 5:56

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) :

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)

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 $$b$$.

• 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 May 19, 2023 at 20:23
• @JacobAvenaim It is good now for you ?
– 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. $$e^{x^{3}}\frac{dy}{dx} +{3x^2}e^{x^{3}}{y}=6x^2e^{x^{3}}$$

to:

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

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