The differential equation, having general solution as $\mathrm{A} x^2+\mathrm{B} y^2=1$, where A and B are…

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. $\quad x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
  2. $x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
  3. $x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
  4. $\quad x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$

Solution

$\mathrm{A} x^2+\mathrm{B} y^2=1$
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 $x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$

Asked in: MHT CET 2024 (15 May Shift 1)

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