Consider the differential equation $y\, dx- (x+y)^2 dy = 0 .$ If for $y = 1, x$ takes value 1, then value of x when $y=4$, is
Answer & explanation
Correct answer: option 3
The correct answer is option (3) : 16
We have$ y\, dx-(x+y^2)dy = 0 $
$⇒\frac{dx}{dy} + \left(-\frac{1}{y}\right) x= y $ ...............(i)
This is a linear differential equation with $I.F.=e^{∫\frac{1}{y}dy}=\frac{1}{y}$
Multiplying both sides of (i) by $I.F=\frac{1}{y}$ and integrating with respect to y, we get
$\frac{x}{y} = ∫y×\frac{1}{y} dy +C$
$⇒\frac{x}{y} = y +C$
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 $.