# Find the general solution of the differential equation? y dy/dx-2e^x=0

Apr 15, 2018

$y = \pm 2 \sqrt{{e}^{x} + A}$

#### Explanation:

We wish to solve the differential equation $y \cdot y ' - 2 {e}^{x} = 0$.

This is a separable, first-order ordinary differential equation. As such, it can be solved using techniques suitable for separable 1st-order ODE's.

The most straightforward technique is to get our equation in the form $f \left(y\right) \mathrm{dy} = g \left(x\right) \mathrm{dx}$. We can then integrate both sides, ridding ourselves of the $y '$ term.

We get our function into this form as such:

$y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {e}^{x} = 0$
$y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 {e}^{x}$
$\left(y\right) \mathrm{dy} = \left(2 {e}^{x}\right) \mathrm{dx}$

See that our left-hand side is a function of just $y$ and our right-hand side a function of just $x$. Integrate both sides.

$\int \left(y\right) \mathrm{dy} = \int \left(2 {e}^{x}\right) \mathrm{dx}$
$\frac{1}{2} {y}^{2} = 2 {e}^{x} + C$
${y}^{2} = 4 {e}^{x} + 2 C$
$y = \pm \sqrt{4 {e}^{x} + 2 C} = \pm 2 \sqrt{{e}^{x} + \frac{C}{2}}$

Since $C$ is an arbitrary constant, let $A = \frac{C}{2}$. Then our final answer is $y = \pm 2 \sqrt{{e}^{x} + A}$.