# What is a solution to the differential equation dy/dx=2yx+yx^2?

Jul 3, 2016

$= \alpha {e}^{{x}^{2} \left(1 + \frac{x}{3}\right)}$

#### Explanation:

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

separate it, starting with....

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

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x + {x}^{2}\right)$

you might recognise $\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}}$ as $\left(\ln y\right) '$ and so the integration can be done pretty infomally, but if not....we can integrate both sides as follows.....starting again with the last equation:

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x + {x}^{2}\right)$

$\frac{1}{y} \mathrm{dy} = \left(2 x + {x}^{2}\right) \setminus \mathrm{dx}$

$\int \mathrm{dy} q \quad \frac{1}{y} = \int \mathrm{dx} q \quad \left(2 x + {x}^{2}\right)$

$\ln y = {x}^{2} + {x}^{3} / 3 + C$

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

$\implies y = {e}^{{x}^{2} \left(1 + \frac{x}{3}\right) + C}$

$= {e}^{{x}^{2} \left(1 + \frac{x}{3}\right)} \cdot {e}^{C}$

$= \alpha {e}^{{x}^{2} \left(1 + \frac{x}{3}\right)}$

where $\alpha = c o n s t = {e}^{C}$

Jul 3, 2016

$y = {C}_{0} {e}^{{x}^{2} + \frac{1}{3} {x}^{3}}$

#### Explanation:

Grouping variables

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

integrating both sides

${\log}_{e} y = {x}^{2} + \frac{1}{3} {x}^{3} + C$

Finally

$y = {C}_{0} {e}^{{x}^{2} + \frac{1}{3} {x}^{3}}$