Question

The curve $f(x,y)=0$ passing through$(0,2)$ satisfy the differential equation $\frac{dy}{dx}=\frac{y^3}{e^x+y^2}.$ If the line $x=\ln 5$ intersects the curve at points $y=\alpha$ and $y=\beta ,$ then $\alpha +\beta$ is

Answer 1

$\frac{dy}{dx}=\frac{y^3}{e^x+y^2}$
$\Rightarrow \frac{dx}{dy}=\frac{e^x}{y^3}+\frac{1}{y}$
$\Rightarrow -y{e}^{-x}.\frac{dx}{dy}=\frac{-1}{y^2}-e^{-x}$
$\Rightarrow e^{-x}-y{e}^{-x}\frac{dx}{dy}=\frac{-1}{y^2}$
$\Rightarrow \frac{d}{dy}\left(y{e}^{-x}\right)=\frac{-1}{y^2}$
$\Rightarrow y{e}^{-x}=\frac{1}{y}+C$
The curve passes through $(0,2).$
$\Rightarrow 2=\frac{1}{2}+C\Rightarrow C=\frac{3}{2}$
$\therefore$ The solution is $y{e}^{-x}=\frac{1}{y}+\frac{3}{2}\cdots \left(1\right)$
Solving $\left(1\right)$ with $x=\ln 5,$ we get
$\frac{1}{5}y=\frac{1}{y}+\frac{3}{2}$
$\Rightarrow 2y^2-15y-10=0$
Given, the solutions are $y=\alpha$ and $y=\beta$
$\therefore \alpha +\beta =\frac{15}{2}=7.50$