Question

$solve \: the \: differential \: equation \:$
$x.cos( \frac{y}{x} )(y.dx + x.dy) = y.cos( \frac{y}{x} )(x.dy + y.dx)$

$all \: steps \: to \: be \: needed \: for \: answer$
☆Challenging question☆​

Answer 1

$\sf x.cos \left( \dfrac{y}{x}\right) (y.dx+x.dy)=y.sin \left( \dfrac{y}{x}\right) (x.dy-y.dx)}$

$\\\to \sf xy.cos\left( \dfrac{y}{x} \right) dx+x^2.cos \left( \dfrac{y}{x}\right) dy=xy.sin\left( \dfrac{y}{x} \right) dy - y^2.sin \left( \dfrac{y}{x}\right) dx$

$\\ \to \sf dy \left[ xy.sin \left(\dfrac{y}{x}\right) - x^2.cos \left(\dfrac{y}{x} \right)\right]=dx \left[ xy.cos \left(\dfrac{y}{x}\right)+ y^2.sin \left(\dfrac{y}{x} \right) \right]$$\\ \to \sf \dfrac{dy}{dx}= \dfrac{ \left[ xy.cos \left(\dfrac{y}{x}\right)+ y^2.sin \left(\dfrac{y}{x} \right) \right]}{\left[ xy.sin \left(\dfrac{y}{x}\right) - x^2.cos \left(\dfrac{y}{x} \right)\right]}\ \; \dashrightarrow\ (1)$

Let , $\sf v=\dfrac{y}{x}\ \; \implies y=vx$

$\sf \bold\red{On differentiating ,}$

$\implies \sf \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

On sub. eqs. in eq. (1) , we get ,

$\\ \to \sf v+x\dfrac{dv}{dx}= \dfrac{ \left[ xy.cos(v)+ y^2.sin(v) \right]}{\left[ xy.sin (v) - x^2.cos(v) \right]}$

Taking 'x²' common in RHS , we get ,

$\\ \to \sf v+x\dfrac{dv}{dx}= \dfrac{ \left[ \dfrac{y}{x}.cos(v)+ \left( \dfrac{y}{x}\right)^2.sin(v) \right]}{\left[ \dfrac{y}{x}.sin (v) -cos(v) \right]}$

Again sub. eq , y = vx , we get ,

$\\ \to \sf v+x\dfrac{dv}{dx}= \dfrac{ \left[v.cos(v)+ (v)^2.sin(v) \right]}{\left[ v.sin (v) -cos(v) \right]}$

$\\ \to \sf v+x\dfrac{dv}{dx}= \dfrac{v\ cos\ v + v^2\ sin\ v}{v\ sin\ v-cos\ v}$

$\\\to \sf x\dfrac{dv}{dx}= \dfrac{v\ cos\ v + v^2\ sin\ v}{v\ sin\ v-cos\ v}-v$

$\\\to \sf x\dfrac{dv}{dx}= \dfrac{2vcos\ v}{v\ sin\ v-cos\ v}$

$\\\to \sf \dfrac{v\ sin\ v-cos\ v}{v\ cos\ v}\ dv= 2\ \dfrac{dx}{x}$

Integrating on both sides ,

$\\\displaystyle \to \int \sf \dfrac{v\ sin\ v-cos\ v}{v\ cos\ v}\ dv=2\ \int \dfrac{dx}{x}$

$\\\displaystyle \to \int \sf \dfrac{cos\ v- v\ sin\ v}{v\ cos\ v}\ dv=-2\ \int \dfrac{dx}{x}$

Use sub. for LHS where , u = v cos v

$\\\to \sf \log (v\ cos\ v)= -2 \log\ x+ \log\ c$

$\\\to \sf \log (v\ cos\ v)+ \log\ x^2= \log$

$\\\to \sf \log (x^2.v\ cos\ v)= \log\ c$

$\\\to \sf x^2.v\ cos\ v= c$

We know that , y = vx

$\\\pink{\leadsto \textsf{\textbf{ xy cos}}\ \dfrac{\textsf{\textbf{y}}}{\textsf{\textbf{x}}} \textsf{\textbf{ = c}}}\ \; \bigstar$