Toronto Math Forum
MAT2442018S => MAT244Tests => Final Exam => Topic started 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.

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}

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.

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 $5y^3$.

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 $5y^3$.

.

I think the solution is wrong, my h(y) was coming out to be just a constant, c

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.

There is a mistake in step (9)
the first term should be 2y(x^2)cosy instead of 2x(y^2)cosy

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