# If the solution curve of the differential equation

Question:

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

1. $y^{5}-2 y-2=0$

2. $2 y^{5}-2 y-1=0$

3. $2 y^{5}-y^{2}-2=0$

4. $y^{5}-y^{2}-1=0$

Correct Option: , 4

Solution:

$\left(2 x-10 y^{3}\right) d y+y d x=0$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}+\left(\frac{2}{\mathrm{y}}\right) \mathrm{x}=10 \mathrm{y}^{2}$

I. F. $=e^{\int_{y}^{2} d y}=e^{2 \ln (y)}=y^{2}$

Solution of D.E. is

$\therefore \quad x \cdot y=\int\left(10 y^{2}\right) y^{2} \cdot d y$

$x y^{2}=\frac{10 y^{5}}{5}+C \Rightarrow x y^{2}=2 y^{5}+C$

It passes through $(0,1) \rightarrow 0=2+\mathrm{C} \Rightarrow \mathrm{C}=-2$

$\therefore$ Curve is $x y^{2}=2 y^{5}-2$

Now, it passes through $(2, \beta)$

$2 \beta^{2}=2 \beta^{5}-2 \Rightarrow \beta^{5}-\beta^{2}-1=0$

$\therefore \beta$ is root of an equation $y^{5}-y^{2}-1=0$