# What are Separable differential equations?

Back when we first learned about derivatives, we used the formal definition to find the derivatives of certain functions. Then we expanded our "derivative taking skills", using more efficient methods such as the power rule, quotient rule and chain rule. Once we mastered taking derivatives of explicitly defined functions (functions in y= form), we moved onto implicitly defined functions. Here we needed to find a way to take the derivative of functions that had the x's and y's mixed together. Like $sin(xy)=x^2+3y$

This same transition also surfaces with integrals. We began by solving indefinite integrals of basic functions. Then we expanded your ability to find anti-derivatives of explicitly defined functions using the methods(integration by parts, partial fraction and u-substitutions). I am sure you can guess what next on the agenda...

How can we find the integral of an expression where the x's and y's and mixed together? One approach is the technique known as separable differential equations. This is a topic that is part of the AP Calculus curriculum and in differential equations course

### What are the steps to solving Separable differential equations?

1. Look at the equation and identify the x's and y's
• Treat $dx$ as an x and $dy$ as a y
2. Separate the the x's and y's,so that they are on opposite sides of the equation.
• It is very important that one sided of the equation is multiplied by dx and the other by dy . In the end dx and dy should never be in the denominator.
3. Integrate both sides
• Once we integrate we only need to add "+C" add to the left side of the equation
4. If possible, get our answer into y=f(x) form
5. Finally, if you are given an initial condition, solve for "+C" by plug in the given x and y value
Note: occasional we will interchange steps 4 and 5.

### An example without an initial condition

$$\frac{dy}{dx}=-8xy^2$$

### An example with an initial condition

Given the initial condition y(0)=4 solve the differential equations: $$(1+x^3)\frac{dy}{dx}=\frac{x^2}{y}$$

### A $2^{nd}$ example with an initial condition

Given the initial condition y(16)=1 solve the differential equations: $$\frac{dy}{dx}-\frac{y}{4\sqrt{x}}=0$$