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

### An example without an initial condition

$$\frac{dy}{dx}=-8xy^2$$
Step 1- Highlight the x's and y's.Separate the x's and y's, treat $dx$ as an $x$ and $dy$ as a $y$. $$\frac{\color{Magenta}{dy}}{\color{cyan}{dx}}=\color{cyan}{-8x}\color{Magenta}{y^2}$$
Step 2- Separate the x's and y's, so that they are on opposite sides of the equation. It is very important that one side of the equation is multiplied by dx and the other by dy.
$$\underbrace{\color{Magenta}{\frac{dy}{y^2}}}_{\color{Magenta}{\text{y part of the equation}}}=\underbrace{\color{Cyan}{-8xdx}}_{\color{Cyan}{\text{x part of the equation}}}$$
Step 3- Integrate both sides (Remember we only need to add one +C)
$$\color{Magenta}{\int\frac{dy}{y^2}}=\color{Cyan}{\int-8xdx}$$
$$\color{Magenta}{\int y^{-2} dy}=\color{Cyan}{-8\int x dx}$$
$$\color{Magenta}{-{y^{-1}}}=\color{Cyan}{-4x^2} +C$$
$${-\frac{1}{\color{Magenta}{y}}}=\color{Cyan}{-4x^2} +C $$
Step 4- If possible, get our answer into y=f(x) form
$$-1=\color{Cyan}{(-4x^2} +C)\color{Magenta}{y} $$
$$-\frac{1}{\color{Cyan}{(-4x^2} +C)}=\color{Magenta}{y} $$
$$\color{Magenta}{y}=-\frac{1}{\color{Cyan}{(-4x^2} +C)}$$
$$\boxed{y=-\frac{1}{-4x^2 +C}}$$

### An example with an initial condition

Given the initial condition y(0)=4 solve the differential equations: $$(1+x^3)\frac{dy}{dx}=\frac{x^2}{y}$$
Step 1- Highlight the x's and y's,treat $dx$ as an $x$ and $dy$ as a $y$ $$\color{cyan}{(1+x^3)}\frac{\color{Magenta}{dy}}{\color{Cyan}{dx}}=\frac{\color{Cyan}{x^2}}{\color{Magenta}{y}}$$
Step 2- Separate the x's and y's, so that they are on opposite sides of the equation. It is very important that one side of the equation is multiplied by dx and the other by dy.
$$\color{Magenta}{y dy}=\color{Cyan}{\frac{x^2}{1+x^3}dx}$$
Step 3- Integrate both sides (Remember we only need to add one +C)
$$\int \color{Magenta}{y dy}=\int \color{Cyan}{\frac{x^2}{1+x^3}dx}$$
Here we have to perform a u-sub on the x-side:

Let $u=1+x^3$ $\frac{du}{dx}=3x^2$ $dx=\frac{du}{3x^2}$
$$\int \color{Magenta}{y dy}= \int \color{Cyan}{\frac{x^2}{u} \frac{du}{3x^2}}$$
$$\int \color{Magenta}{y dy}= \frac{1}{3}\int \color{Cyan}{\frac{1}{u}du}$$
$$\color{Magenta}{\frac{y^2}{2}}=\color{Cyan}{ln|u|+C}$$
$$\color{Magenta}{\frac{y^2}{2}}=\color{Cyan}{ln|1+x^3|+C}$$
Step 4- If possible, get our answer into y=f(x) form
$$\color{Magenta}{y^2}=\color{Cyan}{2(ln|1+x^3|+C)}$$
$$\color{Magenta}{y}=\color{Cyan}{\sqrt{2(ln|1+x^3|+C)}}$$
Step 5- solve for "+C" by plug in the given x and y value
Since x=0 and y=4, we have:
$$ 4=\sqrt{2(ln|1+0^3|+C)}$$
$$4=\sqrt{2(ln|1|+C)}$$
$$ln(1)=0$$ so,
$$4=\sqrt{2C}$$
$$\therefore C=8$$
Making the finally answer:
$$\boxed{y=\sqrt{2(ln|1+x^3|+8}}$$
$$\frac{(x+1)\cancel{(x+2)}}{3\cancel{(x+2)}}$$

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

Given the initial condition y(16)=1 solve the differential equations: $$\frac{dy}{dx}-\frac{y}{4\sqrt{x}}=0$$
Step 1- Highlight the x's and y's,treat $dx$ as an $x$ and $dy$ as a $y$
$$\frac{\color{Magenta}{dy}}{\color{cyan}{dx}}-\frac{\color{Magenta}{y}}{4\color{cyan}{\sqrt{x}}}=0$$
Step 2- Separate the x's and y's, so that they are on opposite sides of the equation. It is very important that one side of the equation is multiplied by dx and the other by dy.
$$\frac{\color{Magenta}{dy}}{\color{cyan}{dx}}=\frac{\color{Magenta}{y}}{4\color{cyan}{\sqrt{x}}}$$
$$\color{Magenta}{\frac{dy}{y}}=\frac{1}{4}\color{Cyan}{\frac{dx}{\sqrt{x}}}$$
$$\color{Magenta}{\frac{1}{y} dy}=\frac{1}{4}\color{Cyan}{x^{-\frac{1}{2}} dx}$$
Step 3- Integrate both sides (Remember we only need to add one +C)
$$\color{Magenta}{ln |y|} = \frac{1}{2}\color{cyan}{\sqrt{x}}+c$$
Here we will solve for "+C", then tranfrom the equation into y=f(x). Thus completing step 5 before step 4.
Step 5- solve for "+C" by plug in the given x and y value
$$ln |1| = \frac{1}{2}\sqrt{16}+c$$
Since $ln(1)=0$
$$0=\frac{1}{2}(4)+c$$
$$0=2+c$$
$$c=-2$$
So now we have:
$$ln |y| = \frac{1}{2}\sqrt{x}-2$$
Step 4- If possible, get our answer into y=f(x) form
In order to get "y" by itself we must exponentiate both sides:
$$e^{ln |y|} = e^{\frac{1}{2}\sqrt{x}-2}$$
$$ y = \pm e^{\frac{1}{2}\sqrt{x}-2}$$
A differential equation **must be continous**, so we will either accept the positive or negative answer. Since we know that (16,1) is a point on the curve, we know we are looking for the positive answer and will reject the negative part of the equation.
Making the finally answer:
$$ y = e^{\frac{1}{2}\sqrt{x}-2}$$

##### How did we know to perform step 5 before step 4 here?