Question

Solve the following Differential Equation:

$x.\:Cos\:(\:\dfrac{y}{x}\:)\:(y.dx + x.dy ) = y.\:Sin\:(\:\dfrac{y}{x}\:)\:(x.dy - y.dx)$

(All the necessary steps to be added)
Chapter: Differential Equations

Answer 1

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

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

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

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 ,

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

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

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

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

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

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

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

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

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

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

We know that , y = vx

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

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