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Q.
Consider the differential equation $\frac{d y}{d x}=\frac{y^3}{2\left(x y^2-x^2\right)}$ such that $y(1)=1$, then the particular solution of the equation is
Differential Equations
Solution:
$\frac{d x}{d y}=\frac{2\left(x y^2-x^2\right)}{y^3} \Rightarrow \frac{1 d x}{x^2 d y}=2\left(\frac{1}{x y}-\frac{1}{y^3}\right)$
$\text { Put, } \frac{1}{x}=t $
$\frac{d t}{d y}=2\left(\frac{1}{y^3}-\frac{t}{y}\right) $
$\frac{d t}{d y}+\frac{2 t}{y}=\frac{2}{y^3} \Rightarrow t \cdot y^2=2 \ln y+\ln c $
$\frac{y^2}{x}=\ln c^2 $
$c y^2=e^{\frac{y^2}{x}}$
$u \operatorname{sing} y(1)=1 \text {, we get } c=e $
$\Rightarrow y^2=e^{\frac{y^2}{x}-1}$