Author Topic: FE-P1  (Read 2053 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2439
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
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

  • Full Member
  • ***
  • Posts: 21
  • Karma: 6
    • View Profile
Re: FE-P1
« Reply #1 on: April 11, 2018, 10:50:49 PM »
First we find the integrating factor.
Note that
\begin{equation}
    M_y=2x\cdot\cos(y) - 2xy\cdot\sin(y) - 2y\cdot\cos(x)
\end{equation}
\begin{equation}
    N_x=4x\cdot\cos(y) - 2xy\cdot\sin(y) - 3y\cdot\cos(x)
\end{equation}
\begin{equation}
    N_x - M_y=2x\cdot\cos(y) -  y\cdot\cos(x)
\end{equation}   
\begin{equation}
   (N_x - M_y)/M=1/y
\end{equation}
Therefore, the integrating factor is only dependent on y.
\begin{equation}
    \ln u= \ln y
\end{equation}
\begin{equation}
    u(y)= y
\end{equation}
Multiply u(y) on both sides of the equation. Then we have
\begin{equation}
    \phi_x=2xy^2\cos(y) - y^3\cos(x)
\end{equation}
\begin{equation}
    \phi=x^2y^2\cos(y)  -y^3\sin(x) +h(y)
\end{equation}
\begin{equation}
    \phi_y=2x^2y\cos(y) -x^2y^2\sin(y) - 3y^2\sin(x) +h^\prime(y)
\end{equation}
By comparison, we get
\begin{equation}
    h^\prime(y)=-5y^4
\end{equation}
\begin{equation}
    h^\prime(y)=-y^5
\end{equation}
Then we have
\begin{equation}
    \phi=x^2y^2\cos(y)-y^3\sin(x) -y^5
\end{equation}
The general solution is
\begin{equation}
\phi=x^2y^2\cos(y)-y^3\sin(x) -y^5=c
\end{equation}
« Last Edit: June 26, 2018, 03:02:26 PM by Victor Ivrii »

Tim Mengzhe Geng

  • Full Member
  • ***
  • Posts: 21
  • Karma: 6
    • View Profile
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

  • Elder Member
  • *****
  • Posts: 91
  • Karma: 36
  • MAT3342018F
    • View Profile
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

  • Full Member
  • ***
  • Posts: 21
  • Karma: 6
    • View Profile
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

  • Elder Member
  • *****
  • Posts: 91
  • Karma: 36
  • MAT3342018F
    • View Profile
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

  • Newbie
  • *
  • Posts: 4
  • Karma: 0
    • View Profile
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

  • Full Member
  • ***
  • Posts: 21
  • Karma: 6
    • View Profile
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

  • Full Member
  • ***
  • Posts: 18
  • Karma: 3
  • mat244h1s-winter2018
    • View Profile
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

  • Full Member
  • ***
  • Posts: 21
  • Karma: 6
    • View Profile
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