MAT244--2018F > Term Test 1

TT1 Problem 1 (morning)

(1/1)

**Victor Ivrii**:

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**:

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.

**Zixuan Wang**:

This is my solution for P1.

I did not see the previous post when I was uploading mine. I am sorry:)).

**Victor Ivrii**:

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

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