← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q6b-i — Step-by-Step Solution

10 marks · Section B

Clairaut's equation · ODEs · asked 5× in 13 yrs · Read the full method →

Question

Consider the differential equation xyp2−(x2+y2−1)p+xy=0xyp^2-(x^2+y^2-1)p+xy=0 where p=dydxp=\dfrac{dy}{dx}. Substituting u=x2u=x^2 and v=y2v=y^2 reduce the equation to Clairaut’s form in terms of u,vu,v and p′=dvdup'=\dfrac{dv}{du}. Hence, or otherwise solve the equation.

Technique

p=xyp′p=\tfrac{x}{y}p' converts the equation to up′2−(u+v−1)p′+v=0up'^2-(u+v-1)p'+v=0; solving for vv gives Clairaut’s form v=up′+f(p′)v=up'+f(p').

Solution

Step 1 — Relate pp and p′p'

With u=x2, v=y2u=x^2,\ v=y^2:

p′=dvdu=dv/dxdu/dx=2y (dy/dx)2x=yx p⟹p=xy p′.p'=\frac{dv}{du}=\frac{dv/dx}{du/dx}=\frac{2y\,(dy/dx)}{2x}=\frac{y}{x}\,p\quad\Longrightarrow\quad p=\frac{x}{y}\,p'.

Step 2 — Substitute and simplify

Insert p=xyp′p=\dfrac{x}{y}p' into xyp2−(x2+y2−1)p+xy=0xyp^2-(x^2+y^2-1)p+xy=0:

xy⋅x2y2p′2−(x2+y2−1)xyp′+xy=0.xy\cdot\frac{x^2}{y^2}p'^2-(x^2+y^2-1)\frac{x}{y}p'+xy=0.

Multiply through by yx\dfrac{y}{x}:

x2p′2−(x2+y2−1)p′+y2=0.x^2p'^2-(x^2+y^2-1)p'+y^2=0.

Now replace x2=u, y2=vx^2=u,\ y^2=v:

u p′2−(u+v−1)p′+v=0.u\,p'^2-(u+v-1)p'+v=0.

Solve for vv:

v(1−p′)=up′−up′2−p′ ⇒ v=p′(p′u−u+1)p′−1=u p′+p′p′−1.v(1-p')=u p'-u p'^2-p'\ \Rightarrow\ v=\frac{p'(p'u-u+1)}{p'-1}=u\,p'+\frac{p'}{p'-1}.

Answer

  v=u p′+p′p′−1  \boxed{\;v=u\,p'+\frac{p'}{p'-1}\;}
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