MAT244-2018S > Final Exam

FE-P1

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Victor Ivrii:
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:
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

Tim Mengzhe Geng:
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:

--- Quote from: 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.

--- End quote ---

It is $-5y^3$.

Tim Mengzhe Geng:

--- Quote from: Meng Wu on April 11, 2018, 10:54:49 PM ---
--- Quote from: 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.

--- End quote ---

It is $-5y^3$.

--- End quote ---
Yes you are right and the solution is corresponding to the case $-5y^3$.