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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
MCQ
A solution of $y = 2x\left( {\frac{{dy}}{{dx}}} \right) + {x^2}{\left( {\frac{{dy}}{{dx}}} \right)^4}$ is
  • A
    $y = 2{C^{\frac{1}{2}}}{x^{\frac{1}{4}}} + C$
  • B
    $y = 2\sqrt C {x^2} + {C^2}$
  • C
    $y = 2\sqrt C \left( {x + 1} \right)$
  • ✓
    $y = 2\sqrt {Cx}  + {C^2}$
✓

Answer

Correct option: D.
$y = 2\sqrt {Cx}  + {C^2}$
d
Writing $\mathrm{P}=\frac{\mathrm{dy}}{\mathrm{dx}}$ and differentiating w.r.t. $\mathrm{x} ;$ we have

$\mathrm{P}=2 \mathrm{P}+2 \mathrm{x} \frac{\mathrm{dP}}{\mathrm{dx}}+2 \mathrm{xP}^{4}+4 \mathrm{P}^{3} \mathrm{x}^{2} \frac{\mathrm{dP}}{\mathrm{dx}}$

$\Rightarrow 0=\mathrm{P}\left(1+2 \mathrm{xP}^{3}\right)+2 \mathrm{x} \frac{\mathrm{dP}}{\mathrm{dx}}\left(1+2 \mathrm{P}^{3} \mathrm{x}\right)$

$\Rightarrow \mathrm{P}+2 \mathrm{x} \frac{\mathrm{dP}}{\mathrm{dx}}=0 \Rightarrow 2 \frac{\mathrm{dP}}{\mathrm{p}}=-\frac{\mathrm{dx}}{\mathrm{x}}$

$\Rightarrow 2 \log \mathrm{P}+\log \mathrm{x}=$ constant

$\Rightarrow \mathrm{P}^{2} \mathrm{x}=\mathrm{C} \Rightarrow \mathrm{P}=\sqrt{\mathrm{C} / \mathrm{x}}$

Substituting this value in the given equation, we

get $y=2 \sqrt{C x}+C^{2}$

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