Question

If $y=f_1\left(x\right)$ and $y=f_2\left(x\right)$ are two solutions of the equation $ydx+dy=-e^x{y}^{2}dy$, where $f_1\left(0\right)=1$ and $f_2\left(0\right)=-1$.

The number of solutions of the equation $f_1\left(x\right).f_2\left(x\right)+x^2=0$ is/are

• A. 0.0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

$\frac{ydx+dy}{e^x{y}^2}=-dy$
$\Rightarrow \frac{e^{-x}}{y}dx+\frac{e^{-x}}{y^2}dy=-dy$
$\Rightarrow -d\left(\frac{e^{-x}}{y}\right)=-dy$
$\therefore \frac{e^{-x}}{y}=y+\lambda$
Given, $f_1\left(0\right)=1$
$\Rightarrow \frac{e^{-0}}{1}=1+\lambda$
$\Rightarrow \lambda =0$
Also, $f_2\left(0\right)=-1$
So, $f_1\left(x\right)=e^{\frac{-x}{2}},$ and $f_2\left(x\right)=-e^{\frac{-x}{2}}$
$f_1\left(x\right).f_2\left(x\right)+x^2=0$
$\Rightarrow e^{\frac{-x}{2}}\times (-e^{\frac{-x}{2}})+x^2=0$
$\Rightarrow -e^{-x}+x^2=0$
$\Rightarrow x^2=e^{-x}$
$\Rightarrow e^x.x^2=1$

So, number of solutions of the equation $f_1\left(x\right).f_2\left(x\right)+x^2=0$ is 1.