Differential equation: dy/dx + 3x^2 y = 6x^2 (2nd stage?)

[roshan2004]

After multiplying the given differential equation by its integrating factor we get the first step,but I simply couldnot understand the second stage,pls explain it to me. The question and the steps are given on the attachment.

 

[Deleted member 4993]

Re: Differential equation

After multiplying the given differential equation by its integrating factor we get the first step,but I simply couldnot understand the second stage,pls explain it to me.
The question and the steps are given on the attachment.

The choice of integrating factor is made in such a way that - you get LHS as a "product function"

Differentiate LHS to see how it is equal to the line above.

 

[soroban]

Hello, roshan2004!

Evidently, you don't understand the purpose of an integrating factor.


$\displaystyle \text{Given: }\:\frac{dy}{dx} + P(x)\!\cdot\!y \:=\:Q(x)$

$\displaystyle \text{The integrating factor is: }\:I \;=\;e^{\int P(x)\,dx}$

$\displaystyle \text{Multiply through by }I\!:\quad \underbrace{I\!\cdot\!\frac{dy}{dx} + I\!\cdot\!P(x)\!\cdot\!y}_{\text{derivative of }I\cdot y} \;=\;I\!\cdot\! Q(x)$

. . $\displaystyle \text{The left side is }always\text{ the derivative of }I\cdot y$

$\displaystyle \text{So we have: }\;\frac{d}{dx}(I\!\cdot\!y) = I\!\cdot\!Q(x)$

. . $\displaystyle \text{Then we integrate both sides . . . and solve for }y.$


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle \text{We have: }\;\frac{dy}{dx} + 3x^2y \;=\;6x^2$

$\displaystyle \text{Then: }\;I \:=\:e^{\int3x^2dx} \:=\:e^{x^3}$

$\displaystyle \text{Multiply through by }I\!:\quad \underbrace{e^{x^3}\frac{dy}{dx} + 3x^2e^{x^3}y}_{\text{derivative of }e^{x^3}y} \;=\;6x^2e^{x^3}$

$\displaystyle \text{So we can write: }\;\frac{d}{dx}\left(e^{x^3}y\right) \;=\;6x^2e^{x^3} \quad\Rightarrow\quad d\left(e^{x^3}y\right) \;=\;6x^2e^{x^3}dx$

$\displaystyle \text{Integrate: }\;\underbrace{\int d\left(e^{x^3}y\right)}_{\Downarrow} \;=\;\underbrace{\int 6x^2e^{x^3}dx}_{\Downarrow}$
. . . . . . . . . . $\displaystyle e^{x^3}y \qquad=\qquad 2e^{x^3} + C$


$\displaystyle \text{Divide by }e^{x^3}\:\quad\boxed{ y \;=\;2 + Ce^{\text{-}x^3}}$