Question

If the solution curve of the differential equation $(2x-10y^3)dy+ydx=0,$ passes through the points $(0,1)$ and $(2,\beta )$, then $\beta$ is a root of the equation

• A. $2y^5-y^2-2=0$
• B. $y^5-y^2-1=0$
• C. $y^5-2y-2=0$
• D. $2y^5-2y-1=0$

Answer 1

The correct option is B $y^5-y^2-1=0$

$2xdy-10y^{3}dy+ydx=0$
$\Rightarrow \frac{dy}{dx}=\frac{y}{10y^3-2x}$
$\Rightarrow \frac{dx}{dy}=\frac{10y^3-2x}{y}$
$\Rightarrow \frac{dx}{dy}+\frac{2x}{y}=10y^2\text{(Linear D.E.)}$

$I.F.=e^{\int \frac{2}{y}dy}=y^2$
$\Rightarrow x{y}^2=\int 10y^{4}dy+C$
$\Rightarrow x{y}^2=\frac{10y^5}{5}+C$
$\Rightarrow x{y}^2=2y^5+C$
It passes through point $(0,1)$
$\Rightarrow C=-2$
$\Rightarrow 2y^5-x{y}^2-2=0\cdots \left(i\right)$
Putting $x=2$ in equation $\left(i\right)$ gives the equation whose root is $\beta$
$i.e.y^5-y^2-1=0$