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Q.
If $y=f(x)(y>0)$ satisfies the differential equation $\frac{x d y}{d x}-y=\frac{x^4}{y}$ such that $y(1)=2$. Find the value of $y(-1)$.
Differential Equations
Solution:
$x d y-y d x=\frac{x^4}{y} d x$
$\frac{y}{x} \cdot \frac{x d y-y d x}{x^2}=x d x $
$\frac{y}{x} d\left(\frac{y}{x}\right)=x d x \Rightarrow \frac{y^2}{x^2}=x^2+2 c$
$\Rightarrow y=\sqrt{x^4+2 x^2} \Rightarrow 4=1+2 c \Rightarrow 2 c=3$
$\therefore y=\sqrt{x^4+3 x^2} \Rightarrow y(-1)=2$