# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 08:39:30 PM

Title: FE-P1
Post by: Victor Ivrii on April 11, 2018, 08:39:30 PM
Find the general solution
\begin{multline*}
\bigl(2xy \cos(y)-y^2\cos(x)\bigr)\,dx  +
\bigl(2x^2\cos(y)-yx^2\sin(y)-3y\sin(x)-5y^3\bigr)\,dy=0\,.
\end{multline*}

Hint. Use the integrating factor.
Title: Re: FE-P1
Post by: Tim Mengzhe Geng on April 11, 2018, 10:50:49 PM
First we find the integrating factor.
Note that

M_y=2x\cdot\cos(y) - 2xy\cdot\sin(y) - 2y\cdot\cos(x)

N_x=4x\cdot\cos(y) - 2xy\cdot\sin(y) - 3y\cdot\cos(x)

N_x - M_y=2x\cdot\cos(y) -  y\cdot\cos(x)

(N_x - M_y)/M=1/y

Therefore, the integrating factor is only dependent on y.

\ln u= \ln y

u(y)= y

Multiply u(y) on both sides of the equation. Then we have

\phi_x=2xy^2\cos(y) - y^3\cos(x)

\phi=x^2y^2\cos(y)  -y^3\sin(x) +h(y)

\phi_y=2x^2y\cos(y) -x^2y^2\sin(y) - 3y^2\sin(x) +h^\prime(y)

By comparison, we get

h^\prime(y)=-5y^4

h^\prime(y)=-y^5

Then we have

\phi=x^2y^2\cos(y)-y^3\sin(x) -y^5

The general solution is

\phi=x^2y^2\cos(y)-y^3\sin(x) -y^5=c
Title: Re: FE-P1
Post by: Tim Mengzhe Geng on April 11, 2018, 10:51:53 PM
Actually Professor I remember on the test the last term is -5y^4 instead of -5y^3. I'm not sure whether I'm correct.
Title: Re: FE-P1
Post by: Meng Wu on April 11, 2018, 10:54:49 PM
Actually Professor I remember on the test the last term is -5y^4 instead of -5y^3. I'm not sure whether I'm correct.

It is $-5y^3$.
Title: Re: FE-P1
Post by: Tim Mengzhe Geng on April 11, 2018, 11:15:27 PM
Actually Professor I remember on the test the last term is -5y^4 instead of -5y^3. I'm not sure whether I'm correct.

It is $-5y^3$.
Yes you are right and the solution is corresponding to the case $-5y^3$.
Title: Re: FE-P1
Post by: Meng Wu on April 11, 2018, 11:17:09 PM
.
Title: Re: FE-P1
Post by: Rasagya Monga on April 12, 2018, 01:28:53 AM
I think the solution is wrong, my h(y) was coming out to be just a constant, c
Title: Re: FE-P1
Post by: Tim Mengzhe Geng on April 12, 2018, 09:56:15 AM
I think the solution is wrong, my h(y) was coming out to be just a constant, c
I checked my solution again and didn't find something wrong. So maybe we should leave it to the Prof.
Title: Re: FE-P1
Post by: Syed Hasnain on April 13, 2018, 01:14:20 PM
There is a  mistake in step (9)
the first term should be 2y(x^2)cosy instead of 2x(y^2)cosy
Title: Re: FE-P1
Post by: Tim Mengzhe Geng on April 14, 2018, 02:40:02 AM
There is a  mistake in step (9)
the first term should be 2y(x^2)cosy instead of 2x(y^2)cosy
Thanks and I've changed it