### Author Topic: TT1 Problem 1 (morning)  (Read 3411 times)

#### Victor Ivrii

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##### TT1 Problem 1 (morning)
« on: October 16, 2018, 05:26:59 AM »
Find integrating factor and then a general solution of ODE
\begin{equation*}
4x^2 y\ln (y)+3xy +\bigl(x^3\ln (y)+ x^3+x^2\bigr)y'=0.
\end{equation*}

Also, find a solution satisfying $y(1)=1$.

#### Yulin WANG

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##### Re: TT1 Problem 1 (morning)
« Reply #1 on: October 16, 2018, 09:21:08 AM »
Let $M(x,y) = 4x^{2}ylny + 3xy$, $N(x,y) = x^{3}lny + x^{3} +x^{2}$

Then $My = 4x^{2}lny + 4x^{2} + 3x$,  $Nx = 3x^{2}lny +3x^{2} +2x$

Since, My $\neq$ Nx, so the equation is not exact.

Since $R = (My - Nx)/N = [(4x^{2}lny + 4x^{2} + 3x) - (3x^{2}lny +3x^{2} +2x)] / (x^{3}lny + x^{3} +x^{2}) = 1/x$

So the integrating factor is $u(x) = e^{\lmoustache Rdx} = e^{\lmoustache(1/x)dx} = e^{lnx} = x$

Then multiply u(x) = x on both sides, then the equation becomes exact.

Let $M' = 4x^{3}ylny + 3x^{2}y, N' = x^{4}lny + x^{4} +x^{3}$

Since $\lmoustache M'dx = x^{4}ylny + x^{3}y + h(y)$,and $\lmoustache N'dy = x^{4}ylny + x^{3}y +g(x)$

So $x^{4}ylny + x^{3}y = c$ is the general soluttion.

Since y(1) = 1, so ln1 + 1 = c, then c = 1.

Therefore, $x^{4}ylny + x^{3}y = 1$ is a solution to the IVP.
« Last Edit: October 16, 2018, 02:29:54 PM by Yulin Wang »

#### Zixuan Wang

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##### Re: TT1 Problem 1 (morning)
« Reply #2 on: October 16, 2018, 09:27:14 AM »
This is my solution for P1.
I did not see the previous post when I was uploading mine. I am sorry:)).
« Last Edit: October 16, 2018, 11:09:32 AM by Zixuan Wang »

#### Victor Ivrii

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##### Re: TT1 Problem 1 (morning)
« Reply #3 on: October 18, 2018, 04:27:21 AM »
Yulin did everything right. You need to clean your post, and escape \ln x resulting in $\ln x$

Zixuan, no reason for you to post