Question

Solve the differential equation \((x\,dy-y\,dx)\,y\sin\!\left(\dfrac{y}{x}\right)=(y\,dx+x\,dy)\,x\cos\!\left(\dfrac{y}{x}\right)\).

Hint:

This is a homogeneous equation; substitute y=vx, separate variables, then integrate tan(v) - 1/v.

Answer 1

Step 1: Recognising a homogeneous equation:
Every term has the same total degree, so this is homogeneous. Substitute \(y=vx\), so \(dy=v\,dx+x\,dv\).

Step 2: Substituting into the equation:
\(x\,dy-y\,dx=x(v\,dx+x\,dv)-vx\,dx=x^{2}dv\). Also \(y\,dx+x\,dy=vx\,dx+x(v\,dx+x\,dv)=2vx\,dx+x^{2}dv\).

Step 3: Rewriting the full equation in terms of v:
\(x^{2}dv\cdot vx\sin v=(2vx\,dx+x^{2}dv)\cdot x\cos v\), i.e. \(v x^{3}\sin v\,dv=(2vx^{2}\cos v)dx+x^{3}\cos v\,dv\).

Step 4: Separating variables:
\(x^{3}(v\sin v-\cos v)dv=2vx^{2}\cos v\,dx\ \Rightarrow\ \dfrac{v\sin v-\cos v}{v\cos v}dv=\dfrac{2\,dx}{x}\ \Rightarrow\ \Big(\tan v-\dfrac1v\Big)dv=\dfrac{2\,dx}{x}\).

Step 5: Integrating both sides:
\(\displaystyle\int\Big(\tan v-\dfrac1v\Big)dv=\int\dfrac{2\,dx}{x}\ \Rightarrow\ -\ln|\cos v|-\ln|v|=2\ln|x|+C_{1}\ \Rightarrow\ \ln|v\cos v|=-2\ln|x|-C_1\).

Final Answer:
So \(v\cos v=\dfrac{C}{x^{2}}\); substituting back \(v=\dfrac yx\) and multiplying by \(x^2\): \(xy\cos\!\Big(\dfrac{y}{x}\Big)=C\).\[ \boxed{xy\cos\!\left(\dfrac{y}{x}\right)=C} \]