# What is a solution to the differential equation dy/dx=y?

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Oct 13, 2016

#### Answer:

$y = C \cdot {e}^{x}$ where $C$ is some constant.

#### Explanation:

If you aren't looking for the general solution, but rather just one solution, then sometimes you can figure it out for simple differential equations like this by thinking for a second about what the differential equation literally means.

$\frac{\mathrm{dy}}{\mathrm{dx}} = y$

We're looking for a function, $y$, which has the property that the derivative of $y$ is equal to $y$ itself.

There's one function which you probably learned previously that has exactly this property:

$y = {e}^{x}$.

The function ${e}^{x}$ is so special precisely because its derivative is also equal to ${e}^{x}$. So $y = {e}^{x}$ is one solution to the differential equation.

If you're also interested in finding all solutions to this DE, (or you're not interested in trial-and-error) then you can solve this DE by separation of variables.

Think of $\mathrm{dy}$ and $\mathrm{dx}$ each as discrete variables. So you could do something like multiply both sides by $\mathrm{dx}$ and end up with:

$\iff \mathrm{dy} = y \mathrm{dx}$

And then divide both sides by $y$:

$\iff \frac{\mathrm{dy}}{y} = \mathrm{dx}$

Now, integrate the left-hand side $\mathrm{dy}$ and the right-hand side $\mathrm{dx}$:

$\iff \int \frac{1}{y} \mathrm{dy} = \int \mathrm{dx}$

$\iff \ln | y | = x + C$

Remember to add the constant of integration, but we only need one.

Raise both sides by $e$ to cancel the $\ln$:

$\iff y = \pm {e}^{x + C}$

Now, pulling the $C$ out front:

$\iff y = \pm C {e}^{x}$

Since $C$ can be either positive or negative, we don't really need the $\pm$:

$\iff y = C {e}^{x}$

So there is our general solution: Any constant multiple of ${e}^{x}$ is a solution to the differential equation, which makes sense.

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