UPSC 2020 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Homogeneous Equations and Reduction · ODEs · Read the full method →

Question

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

Homogeneous ODE; substitution $y=vx$ , then variable separation; integral $\int\tan v\,dv=-\ln\cos v$ .

Observation. The equation is homogeneous of degree 2 in $x,y$ . Put $v=\dfrac{y}{x}$ , i.e. $y=vx$ , so $dy=v\,dx+x\,dv$ .

Expand both sides.

LHS: $x\cos v\,(y\,dx+x\,dy)=xy\cos v\,dx+x^2\cos v\,dy$ .

RHS: $y\sin v\,(x\,dy-y\,dx)=xy\sin v\,dy-y^2\sin v\,dx$ .

Bring everything to one side:

\big(xy\cos v + y^2\sin v\big)dx + \big(x^2\cos v - xy\sin v\big)dy = 0 .

With $y=vx$ : $xy=vx^2$ , $y^2=v^2x^2$ . Coefficients:

P=xy\cos v+y^2\sin v = x^2\big(v\cos v+v^2\sin v\big),

Q=x^2\cos v-xy\sin v = x^2\big(\cos v - v\sin v\big).

So $P\,dx+Q\,dy=0$ becomes (dividing by $x^2$ ):

\big(v\cos v+v^2\sin v\big)dx + \big(\cos v - v\sin v\big)(v\,dx+x\,dv)=0 .

Coefficient of $dx$ :

v\cos v+v^2\sin v + v\cos v - v^2\sin v = 2v\cos v .

Coefficient of $dv$ :

x(\cos v - v\sin v).

Hence

2v\cos v\,dx + x(\cos v - v\sin v)\,dv = 0 .

\frac{2\,dx}{x} = -\frac{\cos v - v\sin v}{v\cos v}\,dv = -\left(\frac{1}{v} - \tan v\right)dv .

2\ln x = -\big(\ln v + \ln\cos v\big) + \text{const} = -\ln(v\cos v) + \ln C .

Thus

\ln\big(x^2 v\cos v\big)=\ln C \quad\Rightarrow\quad x^2\,v\cos v = C .

Restore $v=y/x$ , so $x^2\cdot\dfrac{y}{x}\cos\!\dfrac{y}{x}=C$ :

\boxed{\,x\,y\cos\!\left(\frac{y}{x}\right)=C\,}