Find the particular solution for 2 x y+y^2-2 x^2 d y d x =0 ; y=2… - Vidyadip

MCQ

Find the particular solution for $2 x y+y^2-2 x^2 \frac{d y}{d x}=0 ; y=2$ when $x =1$

✓
$y=\frac{2 x}{1-\log |x|}(x \neq 0, x \neq e)$
B
$y=\frac{3 x}{1-\log |x|}(x \neq 0, x \neq e)$
C
$y=\frac{3 x}{1-\log |x|}(x \neq 0, x \neq e)$
D
$y=\frac{5 x}{1+\log |x|}(x \neq 0, x \neq e)$

✓

Answer

Correct option: A.

$y=\frac{2 x}{1-\log |x|}(x \neq 0, x \neq e)$

Let $y = vx$
$\frac{d y}{d x}=v+x \frac{d}{d x}$
Question becomes $v+x \frac{d y}{d z}=\frac{2 x+x^2}{2}$
$x \frac{d x}{d x}=\frac{2 v+v^2}{2}-t$
$x \frac{d x}{d x}=\frac{2 v+v^2-2}{2}$
$2 \int \frac{d}{x^2}=\int \frac{d x}{z}$
$\frac{-2}{w}=\log x+c$
When $x=1 y =2$ we get
$\frac{-2 x}{y}=\log x+c$
$\frac{-2}{2}=\log 1+c$
$\Rightarrow c=-1$
$\frac{-2 x}{y}=\log x-1$
$y=\frac{2 x}{1-\log x}$

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