Question

3. (20 points) Find the solution to the differential equation y sin(y) dx + x(sin y - ycos y) dy = 0

Differential Equations

Answer 1

y sin(y) dr + (sin(y) – y cos(y)) dy = 0

Is
M dr + N dy = 0
with/ $M=y\sin(y),N=x(\sin(y)-y\cos(y))$

So $M_y=y\cos(y)+\sin(y),N_x=\sin(y)-y\cos(y)$

We have

My - N, -M y cos(y) + sin(y) – sin(y) + y cos(y) -y sin(y) = 2y cosy -y sin(y) = -2 cot(y)

Integrating factor is $\exp\left(\int (-2\cot(y)) \right )=\exp\left(-2\ln(\sin(y)) \right )=\csc^2(y)$

Multiplying by thi/s factor we get

$y\csc(y)\,dx+x(\csc(y)-y\csc(y)\cot(y))\,dy=0$

We can check that this equation is now exact as $\frac{\partial }{\partial y}\left(y\csc(y) \right )=\frac{\partial }{\partial x}\left(x(\csc(y)-y\csc(y)\cot(y)) \right )$

And so the solution is $u(x,y)=C$ such that $u_x(x,y)=y\csc(y)\Rightarrow u(x,y)=xy\cos(y)+g(y)$

Substituting in $u_y(x,y)=x(\csc(y)-y\csc(y)\cot(y))\Rightarrow g'(y)=0$

So our solution is simply ${\color{Red} xy\csc(y)=C}$

$\blacksquare$