Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii 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 $y(1)=1$.

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.

This is my solution for P1.
I did not see the previous post when I was uploading mine. I am sorry:)).

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