Problem 1 (noon)

[Victor Ivrii]

(a) Find integrating factor and then a general solution of ODE
\begin{equation*}
\bigl(2y+y^2\sin(x)\bigr) + \bigl(\sin(2x)+2y\cos(x)\bigr) y'=0
\end{equation*}

(b) Also, find a solution satisfying $y(\dfrac{\pi}{4})=\sqrt{2}$.

[Yuying Chen]

$\text{(a)}\\$
$M=2y+y^{2}\sin x\qquad M_{y}=\frac{\partial}{\partial y}M=2+2y\sin x\\$
$N=\sin 2x+2y\cos x\quad\quad N_{x}=\frac{\partial}{\partial x}N=2\cos 2x-2y\sin x\\$
$\text{Since $M_{y}\neq N_{x}$, the given differential equation is not exact.}\\ $

$R_2=\frac{M_y-N_x}{N}=\frac{2+2y\sin x-2\cos 2x+2y\sin x}{\sin2 x+2y\cos x}=\frac{4\sin x(y+\sin x)}{2\cos x(y+\sin x)}=2\tan x\\$
$\mu=e^{\int R_2dx}=e^{2\int{\tan x}dx}=\sec^{2} x\\$
$(2y\sec^{2} x+y^{2}\sin x\ sec^{2} x)+(\sin 2x\sec^{2} x+2y\cos x\sec^{2} x)=0\\ \\$

$\text{$\exists \psi{(x,y)}$ such that $\psi_{y}=N$}\\$
$\qquad\quad\psi{(x,y)}=\int{\sin 2x\sec^{2} x+2y\cos x\sec^{2} x}dy\\$
$\qquad\qquad\qquad =2\cos 2x\sec^{2} xy+y^2\sec^{2} x\cos x+h(x)\\$  OK. V.I.
$\qquad\quad\psi_{x}=2y\sec^{2} x\\$ ? V.I.
$\qquad\quad h^{\prime}(x)=y^{2}\sin x\sec^{2} x\\$
$\qquad\quad h(x)=y^{2}\sec x\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=2\cos2 x\sec^{2} xy+y^2\sec^{2} x\cos x+y^{2}\sec x=y\sec x(\sin 2x\sec^{2} x+2y)=C\\$

$\text{(b)}\\$   Wrong substitution. V.I.
$\text{Since $y(\frac{\pi}{4})=\sqrt{2}$}\\$
$C=6\sqrt{2}\\$
$\text{Thus,}\\$
$y\sec x(\sin 2x\sec^{2} x+2y)=6\sqrt{2}\\$

Wrong V.I.

[Yiheng Bian]

Hi, I think you have typo in line4, it should be +but you type=