# How do you solve the differential equation dy/dx=6y^2x, where y(1)=1/25 ?

Apr 26, 2018

$\implies y = \frac{1}{28 - 3 {x}^{2}}$

#### Explanation:

We are given:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {y}^{2} x$

Separate the variables:

$\frac{1}{y} ^ 2 \mathrm{dy} = 6 x \mathrm{dx}$

Integrate both sides:

$\int \frac{1}{y} ^ 2 \mathrm{dy} = \int 6 x \mathrm{dx}$

$- \frac{1}{y} = 3 {x}^{2} + C$

$\frac{1}{y} = - 3 {x}^{2} + C$

$y = - \frac{1}{3 {x}^{2} + C}$

where $C$ is an arbitrary constant of integration.

Now solve for $y \left(1\right)$ to find $C$:

$y \left(1\right) = \frac{1}{25} = - \frac{1}{3 {\left(1\right)}^{2} + C}$

$- \frac{1}{25} = \frac{1}{3 + C}$

$3 + C = - 25$

$C = - 28$

Hence, the final solution is:

$y = - \frac{1}{3 {x}^{2} - 28}$

$\implies y = \frac{1}{28 - 3 {x}^{2}}$