0
$\begingroup$

Given $\frac{y}{x}\cos\frac{y}{x}dx-(\frac{x}{y}\sin\frac{y}{x}+\cos\frac{y}{x})dy = 0$

This is a homogeneous differential equation.

Is it valid to substitute $u=\frac{y}{x}$ and $dy=udx+xdu$ into the equation?

If so, after substitution, simplification, and separation of variables I got the following equation $-\log|x|=\log|u|+\int\frac{1}{u}\cot{u}du$

Singling out $\int\frac{1}{u}\cot{u}du$ to be integrated through integration by parts technique I got $\int\frac{1}{u}\cot{u}du = \cot{u}\log|u|+\int \log|u|\csc^2{u}du$

So, is my initial substitution valid? If so, I'm not sure what substitutions to use next in integrating $\int \log|u|\csc^2{u}du$..

$\endgroup$
1
$\begingroup$

Yes, it is valid but the expression can be simplified in an easier way.

Let $u=\frac yx$. Then, $\frac{dy}{dx}=u+x\frac{du}{dx}$ $$\frac{y}{x}\cos\frac{y}{x}dx-\left(\frac{x}{y}\sin\frac{y}{x}+\cos\frac{y}{x}\right)dy = 0$$ $$\frac{dy}{dx}=\frac{\frac{y}{x}\cos\frac{y}{x}}{\frac{x}{y}\sin\frac{y}{x}+\cos\frac{y}{x}}$$ $$u+x\frac{du}{dx}=\frac{u\cos u}{\frac{1}{u}\sin u+\cos u}$$ $$u+x\frac{du}{dx}=\frac{u^2\cos u}{\sin u+u\cos u}$$ $$x\frac{du}{dx}=\frac{u^2\cos u-u\sin u-u^2\cos u}{\sin u+u\cos u}$$ $$x\frac{du}{dx}=\frac{-u\sin u}{\sin u+u\cos u}$$ $$\int\frac{(\sin u+u\cos u)du}{u\sin u}=\int\frac{-dx}{x}$$ $$\int\left(\frac1u+\cot u\right)du=\int\frac{-dx}{x}$$ $$\ln|u|+\ln|\sin u|=-\ln|x|+c$$

$\endgroup$
  • 1
    $\begingroup$ You took the minus sign in 2 places accidentally $\endgroup$ – Nikunj Mar 12 '16 at 20:26
  • $\begingroup$ @Nikunj Thank you, edited the answer. $\endgroup$ – GoodDeeds Mar 12 '16 at 20:29
1
$\begingroup$

First write $$\frac{dy}{dx}=\frac{\frac{y}{x}\cos {\frac{y}{x}}}{\frac{x}{y}\sin\frac{y}{x}+\cos\frac{y}{x}}$$ Then put $y=vx$ to get, $\frac {dy}{dx}= v+ x\frac{dv}{dx}$ $$\implies x\frac{dv}{dx}=\frac{v^2 \cos v}{\sin v+v\cos v}-v$$ $$\implies \int \frac{(\sin v+v \cos v) dv}{v\sin v}= -\int \frac{dx}{x}$$ $$\implies \ln |v| +\ln |\sin v|= -\ln x+C$$ Now simplify and substitute $v=\frac{y}{x}$, to get the answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.