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)

3

Answer 2

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