Question

The solution of the differential equation $ydx-(x+2y^2)dy=0$ is $x=f(y)$. If $f(-1)=1$, then $f(1)$ is equal to :

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (B)

Given differential equation is
$ydx-\left(x+2y^2\right)dy=0$...(1)
and solution of $(1)$ is $x=f(v)$; where $f(-1)=1,f(1)=?$
Rearranging (1), we get
$y\frac{dx}{dy}-\left(x+2y^2\right)=0$
$\Rightarrow \frac{dx}{dy}-2y-\frac{x}{y}=0$
or $\frac{dx}{dy}+\left(\frac{-1}{y}\right)x=2y$,
which is a linear differential equation of first order $\frac{dx}{dy}+P$
$x=Q$; Its I.F. $=e^{\int Pdy}=e^{\int \frac{-1}{y}dy}=e^{-\ln y}=\frac{1}{y}$
$\therefore$ Solution of (1) is given by
$x.(I.F)=\int Q(I.F.)dy+C$
$\Rightarrow x\cdot \frac{1}{y}=\int 2y\cdot \frac{1}{y}dy+C$
$\Rightarrow \frac{x}{y}=2y+c\Rightarrow x=2y^2+cy;f(-1)=1$
$x+1=2+c(-1)\Rightarrow c=1\therefore x=2y^2+y=f(y)$
$\Rightarrow f(1)=2+1=3$