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MHT CET · Maths · Differential Equations

The differential equation, having general solution as \(\mathrm{A} x^2+\mathrm{B} y^2=1\), where A and B are arbitrary constants, is

  1. A \(x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0\)
  2. B \(x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0\)
  3. C \(x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0\)
  4. D \(x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0\)

Step-by-step Solution

Detailed explanation

\(\mathrm{A} x^2+\mathrm{B} y^2=1\)
Differentiating w.r.t. \(x\), we get
\(2 \mathrm{~A} x+2 \mathrm{~B} y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0...(i)\)
Again, differentiating w.r.t. \(x\), we get
\(2 \mathrm{~A}+2 \mathrm{~B}\left[\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right]=0...(ii)\)
Solving (i) and (ii), we get
\(x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0\)