roshan2004
New member
- Joined
- Oct 29, 2008
- Messages
- 1
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Differential equation: dy/dx + 3x^2 y = 6x^2 (2nd stage?)
- Thread starter roshan2004
- Start date
![differential[1].JPG](/forum/data/attachments/0/198-d7bdb7e2c3fbc3200118a76774fbeade.jpg)
D
Deleted member 4993
Guest
Re: Differential equation
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.
roshan2004 said: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.
Hello, roshan2004!
Evidently, you don't understand the purpose of an integrating factor.
\(\displaystyle \text{Given: }\:\frac{dy}{dx} + P(x)\!\cdot\!y \:=\:Q(x)\)
\(\displaystyle \text{The integrating factor is: }\:I \;=\;e^{\int P(x)\,dx}\)
\(\displaystyle \text{Multiply through by }I\!:\quad \underbrace{I\!\cdot\!\frac{dy}{dx} + I\!\cdot\!P(x)\!\cdot\!y}_{\text{derivative of }I\cdot y} \;=\;I\!\cdot\! Q(x)\)
. . \(\displaystyle \text{The left side is }always\text{ the derivative of }I\cdot y\)
\(\displaystyle \text{So we have: }\;\frac{d}{dx}(I\!\cdot\!y) = I\!\cdot\!Q(x)\)
. . \(\displaystyle \text{Then we integrate both sides . . . and solve for }y.\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
\(\displaystyle \text{We have: }\;\frac{dy}{dx} + 3x^2y \;=\;6x^2\)
\(\displaystyle \text{Then: }\;I \:=\:e^{\int3x^2dx} \:=\:e^{x^3}\)
\(\displaystyle \text{Multiply through by }I\!:\quad \underbrace{e^{x^3}\frac{dy}{dx} + 3x^2e^{x^3}y}_{\text{derivative of }e^{x^3}y} \;=\;6x^2e^{x^3}\)
\(\displaystyle \text{So we can write: }\;\frac{d}{dx}\left(e^{x^3}y\right) \;=\;6x^2e^{x^3} \quad\Rightarrow\quad d\left(e^{x^3}y\right) \;=\;6x^2e^{x^3}dx\)
\(\displaystyle \text{Integrate: }\;\underbrace{\int d\left(e^{x^3}y\right)}_{\Downarrow} \;=\;\underbrace{\int 6x^2e^{x^3}dx}_{\Downarrow}\)
. . . . . . . . . . \(\displaystyle e^{x^3}y \qquad=\qquad 2e^{x^3} + C\)
\(\displaystyle \text{Divide by }e^{x^3}\:\quad\boxed{ y \;=\;2 + Ce^{\text{-}x^3}}\)
Evidently, you don't understand the purpose of an integrating factor.
\(\displaystyle \text{Given: }\:\frac{dy}{dx} + P(x)\!\cdot\!y \:=\:Q(x)\)
\(\displaystyle \text{The integrating factor is: }\:I \;=\;e^{\int P(x)\,dx}\)
\(\displaystyle \text{Multiply through by }I\!:\quad \underbrace{I\!\cdot\!\frac{dy}{dx} + I\!\cdot\!P(x)\!\cdot\!y}_{\text{derivative of }I\cdot y} \;=\;I\!\cdot\! Q(x)\)
. . \(\displaystyle \text{The left side is }always\text{ the derivative of }I\cdot y\)
\(\displaystyle \text{So we have: }\;\frac{d}{dx}(I\!\cdot\!y) = I\!\cdot\!Q(x)\)
. . \(\displaystyle \text{Then we integrate both sides . . . and solve for }y.\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
\(\displaystyle \text{We have: }\;\frac{dy}{dx} + 3x^2y \;=\;6x^2\)
\(\displaystyle \text{Then: }\;I \:=\:e^{\int3x^2dx} \:=\:e^{x^3}\)
\(\displaystyle \text{Multiply through by }I\!:\quad \underbrace{e^{x^3}\frac{dy}{dx} + 3x^2e^{x^3}y}_{\text{derivative of }e^{x^3}y} \;=\;6x^2e^{x^3}\)
\(\displaystyle \text{So we can write: }\;\frac{d}{dx}\left(e^{x^3}y\right) \;=\;6x^2e^{x^3} \quad\Rightarrow\quad d\left(e^{x^3}y\right) \;=\;6x^2e^{x^3}dx\)
\(\displaystyle \text{Integrate: }\;\underbrace{\int d\left(e^{x^3}y\right)}_{\Downarrow} \;=\;\underbrace{\int 6x^2e^{x^3}dx}_{\Downarrow}\)
. . . . . . . . . . \(\displaystyle e^{x^3}y \qquad=\qquad 2e^{x^3} + C\)
\(\displaystyle \text{Divide by }e^{x^3}\:\quad\boxed{ y \;=\;2 + Ce^{\text{-}x^3}}\)