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Let the solution curve of the differential equation
$\mathrm{x \frac{d y}{d x}-y=\sqrt{y^{2}+16 x^{2}}, y(1)=3\: be\: y=y(x)}$ . Then $\mathrm{y(2)}$ is equal to :
Option: 1
$15$

Option: 2
$11$

Option: 3
$13$

Option: 4
$17$

Answers (1)

best_answer

$\mathrm{x \frac{d y}{d x}-y =\sqrt{y^{2}+16 x^{2}} }\\$

$\mathrm{\Rightarrow \frac{d y}{d x} =\frac{y+\sqrt{y^{2}+16 x^{2}}}{x} }$

Let $\mathrm{y=v x} \\$

$\mathrm{\frac{d y}{d x}=v+x \frac{d v}{d x}}$

$\mathrm{\Rightarrow v+x \frac{d v}{d x}=\frac{v x+\sqrt{v^{2} x^{2}+16 x^{2}}}{x} }\\$

$\mathrm{\Rightarrow v+x \frac{d v}{d x}=v+\sqrt{v^{2}+16}} \\$

$\mathrm{\Rightarrow \frac{d v}{\sqrt{v^{2}+16}}=\frac{d x}{x}}$

Integrating

$\mathrm{\Rightarrow \ln \left|v+\sqrt{v^{2}+16}\right|=\ln x+\ln k} \\$

$\mathrm{\Rightarrow v+\sqrt{v^{2}+16}=k x }\\$

$\mathrm{\Rightarrow \frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}+16}=k x}$

$\mathrm{\text { As } y(1)=3} \\$

$\mathrm{\Rightarrow 3+\sqrt{9+16}=k} \\$

$\mathrm{\Rightarrow k=8 }$

$\mathrm{For\: x=2}\\$

$\mathrm{\frac{y}{2}+\sqrt{\frac{y^{2}}{4}+16}=8 \times 2} \\$

$\mathrm{\Rightarrow y+\sqrt{y^{2}+64}=32} \\$

$\mathrm{\Rightarrow \sqrt{y^{2}+64}=-y+32} \\$

$\mathrm{\Rightarrow y^{2}+64=y^{2}+1024-64 y} \\$

$\mathrm{\Rightarrow 64 y=960} \\$

$\mathrm{\Rightarrow y=15}$

Hence answer is option 1

Posted by

manish painkra

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