The substitution \( y=vx \) transforms the differential equation \( xdy-ydx=\sqrt{x^2+y^2}dx \) into
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x} \)
Let y = vx, so dy = v dx + x dv.
Left side:
\( \displaystyle x\,dy-y\,dx=x(v\,dx+x\,dv)-vx\,dx \)
\( \displaystyle =x^2\,dv \)
Right side (taking x > 0):
\( \displaystyle \sqrt{x^2+v^2x^2}\,dx=x\sqrt{1+v^2}\,dx \)
Equating and dividing by \( x^2\sqrt{1+v^2} \):
\( \displaystyle x^2\,dv=x\sqrt{1+v^2}\,dx \)
\( \displaystyle \ \Rightarrow\ \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x} \)
NCERT Exemplar:
Differential Equations
