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If $\mathrm{y= y\left ( x \right )}\\$ is the solution of the dieffrential equation $\mathrm {2x^{2}\frac{dy}{dx}-2xy+3y^{2}=0}\\$ such that $\mathrm{y\left ( e \right )= \frac{e}{3}}\\$ , then $\mathrm{y\left ( 1 \right )}$ is equal to
Option: 1
$\mathrm{\frac{1}{3}}$

Option: 2
$\mathrm{\frac{2}{3}}$

Option: 3
$\mathrm{\frac{3}{2}}$

Option: 4
$\mathrm{3}$

Answers (1)

best_answer

$\mathrm{2x^{2}\frac{dy}{dx}-2xy+3y^{2}= 0}\\$

$\mathrm{\frac{1}{y^{2}}\frac{dy}{dx}-\frac{1}{xy}+\frac{3}{2x^{2}}=0\\ \hspace{0.5cm}Let \hspace{0.2cm} \frac{1}{y}=t}\\$

$\mathrm{-\frac{1}{y^{2}}\frac{dy}{dx}=\frac{dt}{dx}}\\$

$\mathrm{-\frac{dt}{dx}-\frac{1}{x}t+\frac{3}{2x^{2}}=0}\\$

$\mathrm{\frac{dt}{dx}+\frac{1}{x}t=\frac{3}{2x^{2}}}\\$

If $\mathrm{=e^{\int \frac{1}{x}dx}=x}\\$

$\mathrm{x\cdot t=\int x\cdot \frac{3}{2x^{2}}dx}\\$

$\mathrm{x\cdot t=\frac{3}{2}\left ( \ln x \right )+c}\\$

$\mathrm{\frac{x}{y}=\frac{3}{2}\ln x+c}\\$

$\mathrm{x=e,y=\frac{e}{3}}\\$

$\mathrm{\frac{e}{e/3}=\frac{3}{2}\cdot 1+c}\\$

$\mathrm{\Rightarrow c=\frac{3}{2}}\\$

$\mathrm{\frac{x}{y}=\frac{3}{2}\ln x+\frac{3}{2}}\\$

$\mathrm{\Rightarrow x=1}\\$

$\mathrm{\Rightarrow \frac{1}{y}=\frac{2}{3}}\\$

$\mathrm{y=\frac{2}{3}}\\$ ........... $\mathrm{\because y=y(x)}\\$

$\mathrm{y=y(1)}$

Hence the correct answer is option (2)

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