Question

The solution of $(2x - 10y^3) \frac{dy}{dx} + y= 0$ is

• A. $xy^2=2y^5 +C$
• B. $yx^2 = 2y^5 + C$
• C. $x^2y^2 = 2y^5 + C$
• D. $None\, of\, these$

Answer 1

Answer: (A)

Given differential equation is $\left(2x - 10y^{3}\right) \frac{dy}{dx} +y= 0 $
or $\frac{dy}{dx} = \frac{-y}{2 x + 10y^3} $
or $\frac{dy}{dx} = \frac{2x - 10y^3}{-y} = \frac{-2x}{y} + 10 y^2$
or $\frac{dy}{dx} + \frac{2}{y} x = 10 y^2$
Compare with linear differential equation
$ \frac{dx}{dy} + Px =Q$
I.F $= e^{\int \frac{2}{y} dy} =e^{\log y^2} =y^{2}$
$ \therefore $ Required solution is
$x.y^2 \int 10y62 .y^2 \, dy +C$
or $x \times y^2 = 10 \times \frac{y^5}{5} + C$
or $xy^2 = 2y^5 + C$