### Author Topic: FE-P1  (Read 5677 times)

#### Victor Ivrii

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##### FE-P1
« 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.

#### Tim Mengzhe Geng

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##### Re: FE-P1
« Reply #1 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
« Last Edit: June 26, 2018, 03:02:26 PM by Victor Ivrii »

#### Tim Mengzhe Geng

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##### Re: FE-P1
« Reply #2 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.

#### Meng Wu

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##### Re: FE-P1
« Reply #3 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 $-5y^3$.

#### Tim Mengzhe Geng

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##### Re: FE-P1
« Reply #4 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 $-5y^3$.

#### Meng Wu

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##### Re: FE-P1
« Reply #5 on: April 11, 2018, 11:17:09 PM »
.
« Last Edit: April 13, 2018, 01:57:33 PM by Meng Wu »

#### Rasagya Monga

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##### Re: FE-P1
« Reply #6 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

#### Tim Mengzhe Geng

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##### Re: FE-P1
« Reply #7 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.

#### Syed Hasnain

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##### Re: FE-P1
« Reply #8 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

#### Tim Mengzhe Geng

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##### Re: FE-P1
« Reply #9 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