MAT244--2018F > Final Exam

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Victor Ivrii:

Find the general solution
\begin{equation*}
\bigl[2x\sin(y) +1\bigr]\,dx  +
\bigl[4x^2\cos(y) + 3x\cot(y)+5 \sin(2y)\bigr]\,dy=0\,.
\end{equation*}
Hint: Use the integrating factor.

Jingze Wang:
Let $M=2x\sin(y)+1, N=4x^2\cos(y)+3x\cot(y)+5\sin(2y)$
$M_y=2x\cos(y), N_x=8x\cos(y)+3\cot(y)$
Check and find this is not exact
Then try to find integrating factor
$N_x-M_y=8x\cos(y)+3\cot(y)-2x\cos(y)=6x\cos(y)+3\cot(y)$
By observation, $\frac{N_x-M_y}{M}=3\cot(y)$
Therefore, the integrating factor is $\sin^{3}(y)$
$M'=2x\sin^4(y)+\sin^3(y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+h(y)$
$\psi_y=4x^2\sin^3(y)\cos(y) + x \sin^2(y)\cos(y)+h'(y)$
$h'(y)=10\sin^4(y) \cos(y)$
$h(y)=2\sin^5 (y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$

Jingyi Wang:
Do not forget the function of Y. We also have to integrate N.
𝑥2sin4𝑦+𝑥sin3𝑦+2sin5y

Jingze Wang:
Sorry I have not finished my typed solution at that time, so you just saw part of my solution, but we get the same answer finally :)

Jingyi Wang:
Sorry about that. I did not see that part. Please ignore my post.

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