Consider the differential equation $y d x-\left(x+y^2\right) d y=0$. If for $y=1, x$ takes value 1 , then value of $x$ when $y=4$, is
Answer & explanation
Correct answer: option 3
We have, $y d x-\left(x+y^2\right) d y=0$
$\Rightarrow \frac{d x}{d y}+\left(-\frac{1}{y}\right) x=y$ ....(i)
This is a linear differential equation with integrating factor = $e^{\int \frac{1}{y} d y}=\frac{1}{y}$
Multiplying both sides of (i) by integrating factor = $\frac{1}{y}$ and integrating with respect to $y$, we get
$\frac{x}{y}=\int y \times \frac{1}{y} d y+C$
$\Rightarrow \frac{x}{y}=y+C$ .....(ii)
It is given that $y=1$ when $x=1$. Putting $x=1, y=1$ in (ii), we get $C=0$. Putting $C=0$ in (ii), we get $x=y^2$. When $y=4$ this equation gives x = 16.