Solve ODE reducible to exact equation

[mahler1]

Homework Statement .

Find all the solutions of the equation:
$\dfrac{\sin(y)}{x}dx+(\dfrac{y}{x}\cos(y)-\dfrac{\sin(y)}{y})dy=0$ knowing that the equation admits an integrating factor $u$ of the form $u(x,y)=h(\dfrac{x}{y})$

The attempt at a solution.

If I call $M(x,y)=\dfrac{\sin(y)}{x}$ and $N(x,y)=\dfrac{y}{x}\cos(y)-\dfrac{\sin(y)}{y}$ then, in order to be an exact differential equation, $M$ and $N$ must satisfy $M_y=N_x$, but $M_y=\dfrac{\cos(y)}{x}≠-\dfrac{y\cos(y)}{x^2}=N_x$, so this is not an exact differential equation.

As the exercise suggests, I've tried to propose an integrating factor $u$ of the form $u(x,y)=h(\dfrac{x}{y})$, but, since up to know I've only solved equations where the integrating factor depended just on one of the variables, I got stuck in the middle of the problem.

So, $u$ must satisfy:

$(uM)dx+(uN)dy=0$

Then, $(uM)_y=(uN)_x \iff u_yM+uM_y=u_xN+uN_x \iff xh_yM+hM_y=\dfrac{1}{y}h_xN+hN_x$.

Calculating all these partial derivatives gives:

$xh_y\dfrac{\sin(y)}{x}+h\dfrac{\cos(y)}{x}=\dfrac{1}{y}h_x(\dfrac{ycos(y)}{x}-\dfrac{\sin(y)}{y})+h(-\dfrac{ycos(y)}{x^2})$

At this point I got stuck and I don't know what and how to integrate in order to find $u(x,y)$

 

[vela]

You didn't differentiate correctly. You should have, for example,
\begin{align*}
(uM)_y &= \frac{\partial}{\partial y} \left[ h\left(\frac{x}{y}\right) \frac{\sin y}{x}\right] \\
&= h'\left(\frac{x}{y}\right)\left(-\frac{x}{y^2}\right) \frac{\sin y}{x} + h\left(\frac{x}{y}\right) \frac{\cos y}{x} \\
&= -h'\left(\frac{x}{y}\right) \frac{\sin y}{y^2} + h\left(\frac{x}{y}\right) \frac{\cos y}{x}
\end{align*}

 

[mahler1]

You didn't differentiate correctly. You should have, for example,
\begin{align*}
(uM)_y &= \frac{\partial}{\partial y} \left[ h\left(\frac{x}{y}\right) \frac{\sin y}{x}\right] \\
&= h'\left(\frac{x}{y}\right)\left(-\frac{x}{y^2}\right) \frac{\sin y}{x} + h\left(\frac{x}{y}\right) \frac{\cos y}{x} \\
&= -h'\left(\frac{x}{y}\right) \frac{\sin y}{y^2} + h\left(\frac{x}{y}\right) \frac{\cos y}{x}
\end{align*}

Thanks, I'll correct it and see if I can get a nicer expression.

 

[mahler1]

You didn't differentiate correctly. You should have, for example,
\begin{align*}
(uM)_y &= \frac{\partial}{\partial y} \left[ h\left(\frac{x}{y}\right) \frac{\sin y}{x}\right] \\
&= h'\left(\frac{x}{y}\right)\left(-\frac{x}{y^2}\right) \frac{\sin y}{x} + h\left(\frac{x}{y}\right) \frac{\cos y}{x} \\
&= -h'\left(\frac{x}{y}\right) \frac{\sin y}{y^2} + h\left(\frac{x}{y}\right) \frac{\cos y}{x}
\end{align*}

After vela's correction, I got:

$(uM)_y=-h'(\dfrac{x}{y})(\dfrac{\sin(y)}{y^2})+h(\dfrac{x}{y})(\dfrac{\cos(y)}{x})$.
$(uN)_x=h'(\dfrac{x}{y})(\dfrac{\cos(y)}{x}-\dfrac{\sin(y)}{y^2})-h(\dfrac{x}{y})\dfrac{y\cos(y)}{x^2})$.
So, $(uM)_y=(uN)_x$ if and only if

$-h'(\dfrac{x}{y})\dfrac{\sin(y)}{y^2}+h(\dfrac{x}{y})\dfrac{\cos(y)}{x}=h'(\dfrac{x}{y})(\dfrac{\cos(y)}{x}-\dfrac{\sin(y)}{y^2})-h(\dfrac{x}{y})\dfrac{y\cos(y)}{x^2})$ if and only if

$h(\dfrac{x}{y})\dfrac{\cos(y)}{x}=h'(\dfrac{x}{y})(\dfrac{\cos(y)}{x})-h(\dfrac{x}{y})\dfrac{y\cos(y)}{x^2})$.

Then

$h(\dfrac{x}{y})(\dfrac{\cos(y)}{x}+\dfrac{y\cos(y)}{x^2})=h'(\dfrac{x}{y})(\dfrac{\cos(y)}{x})$

$1+\dfrac{y}{x}=\dfrac{h'\dfrac{x}{y}}{h\dfrac{x}{y}}$. If I make the substitution $z=\dfrac{x}{y}$, then the equation reduces to:

$1+\dfrac{1}{z}=\dfrac{h'(z)}{h(z)}$.

So

$\int (1+\dfrac{1}{z})dz=\int \dfrac{dh}{h} \iff z+ln(z)+k=ln(h)$.

Then, $h=ze^z=\dfrac{x}{y}e^{\dfrac{x}{y}}$. Now I have to go back to the original equation, which it will be reduced to an exact equation by this integrating factor.

Is this correct?

 

[vela]

Yes, it is!

 

[mahler1]

Yes, it is!

Thanks very much!

 

[lurflurf]

Good work.
Let me offer a suggestion.
I would change variable z=x/y first to make things less messy
$$\frac{\sin(y)}{x}\mathrm{d}x+\left( \frac{y}{x}\cos(y)-\frac{\sin(y)}{y}\right)\mathrm{d}y=0\\
\text{becomes}\\
\sin(y)\frac{\mathrm{d}z}{z}+\frac{1}{z}\cos(y)\mathrm{d}y=0$$

 

[mahler1]

Good work.
Let me offer a suggestion.
I would change variable z=x/y first to make things less messy
$$\frac{\sin(y)}{x}\mathrm{d}x+\left( \frac{y}{x}\cos(y)-\frac{\sin(y)}{y}\right)\mathrm{d}y=0\\
\text{becomes}\\
\sin(y)\frac{\mathrm{d}z}{z}+\frac{1}{z}\cos(y)\mathrm{d}y=0$$

Good remark.