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Separation Of Variables 

classification: this technique applies to firstorder, separable differential equations 
An equation that allows us to separate the variables is called a separable differential equation. 
This is one of the simplest techniques when solving differential equations and sometimes it is introduced in first year calculus while studying integration. This technique is used when it is possible to isolate the variables, using algebra and differentials, on separate sides of the equal sign. This allows us to integrate both sides of the equation individually with only one variable on each side.
The idea with this technique is that the differential equation is in a form where we can isolate the two variables to each side of the equal sign. Then we can integrate each side separately. It's really that easy. The trick is to use algebra to get the equation into the right form.
Separable Equation Form 

Some textbooks write separable equations as \(\displaystyle{ N(y)\frac{dy}{dx}=M(x) }\). This notation tells you that \(N(y)\) is some function that contains only \(y\)'s and maybe some constants. Similarly, \(M(x)\) is some function that contains only \(x\)'s and maybe some constants.
A better way to write a separable equation is an equation of the form \(\displaystyle{ \frac{dy}{dx}=g(x)h(y) }\). In this equation, it is easy to see what needs to done. We need to have two functions multipled together, with one having only x's and the other with only y's.
As we work this example, we will give you a heads up on something. Some instructors (most, from what I've seen), do a trick with constants that confuse a lot of people. Let's do an example to demonstrate the separation of variables technique.
Example 
Find y for \( y' = x \). 

For this separable equation, using the form \(\displaystyle{ \frac{dy}{dx}=g(x)h(y) }\), we have \(g(x)=x\) and \(h(y)=1\)
\(\displaystyle{ y' = x ~~~ \to ~~~ \frac{dy}{dx} = x }\) 
First, we rewrite the derivative so that it is more obvious what we need to do. 
\(\displaystyle{ \frac{dy}{dx} = x ~~~ \to ~~~ dy = x~dx}\) 
Now we write the differential equation by 'moving' the \(dx\) to the other side. It looks like we are multiplying \(dx\) on both sides but that's not what is really happening. However, it helps me remember what to do by thinking of it this way. What we are doing is writing the equation in differential form. 
\(\displaystyle{ \begin{array}{rcl} \int{dy} & = & \int{x~dx} \\ y + c_1 & = & \frac{x^2}{2} + c_2 \end{array} }\) 
Next, we integrate both sides. On the left we integrate with respect to y, on the right with respect to x. 
\(\displaystyle{ y = \frac{x^2}{2} + c_2c_1 }\) 
Okay, so now we have two constants, one from each integration (don't forget these constants; they are extremely important when solving differential equations). We are going to move \(c_1\) to the other side and combine it with \(c_2\). 
\(\displaystyle{ y = \frac{x^2}{2} + C }\) 
Let \( C = c_2  c_1 \) to get our final answer. 
Notes
1. This is a general solution. To get the particular solution, we need initial conditions to determine the value of C.
2. Many instructors skip the steps showing the two constants and combining them, i.e. \( C = c_2  c_1 \) . They just jump to one constant. This can be confusing at first, if you don't know what is happening. Here is a video explaining this in more detail.
Krista King Math  How can the constant C absorb all that other stuff? [3min54secs]  
And, that's it! That's all there is to it. Just separate the variables and integrate. There is nothing fancy going on and no tricks. But this is a basic technique that you need to understand how to do and practice thoroughly since, later on in your study of differential equations, you will learn other techniques to convert more complicated equations into this form.
Here is an indepth video discussing firstorder linear equations, separation of variables and steadystate and transient solutions. If you are first starting to learn differential equations, this may be a bit above you but you can still get a lot out of it and being exposed to a little more advanced techniques will help you learn them later.
MIT OCW  Solving Firstorder Linear ODE's; Steadystate and Transient Solutions.  
After you work through the practice problems, 
firstorder, linear differential equations → 
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Practice Problems 

Instructions   Unless otherwise instructed, solve these separable differential equations and find the particular solution, if possible. Give your answers in exact terms.
Level A  Basic 
Practice A01  

\(\displaystyle{\frac{dy}{dx}=xy}\)  
solution 
Practice A02  

\(\displaystyle{\frac{dy}{dx}=y^2(1+x^2)}\); \(y(0)=1\)  
solution 
Practice A03  

\(\displaystyle{\frac{dy}{dx}=\cos(x)}\); \(y(0)=1\)  
solution 
Practice A04  

\(\displaystyle{\frac{dy}{dx}=x/y^2}\)  
solution 
Practice A05  

\(\displaystyle{\frac{dy}{dx}\cdot\frac{y^3+y}{x^2+3x}=1}\)  
solution 
Practice A06  

\(\displaystyle{\frac{dy}{dx}=\frac{\cos(x)}{y1}}\)  
solution 
Practice A07  

\(\displaystyle{\frac{dy}{dx}=e^{4xy}}\); \(y(0)=5\)  
solution 
Practice A08  

\(\displaystyle{\frac{dy}{dx}=2x\sqrt{y1}}\)  
solution 
Practice A09  

\(\displaystyle{\frac{y+2}{x^2x+2}\frac{dy}{dx}=\frac{x}{y}}\); \(y(1)=2\)  
solution 
Practice A10  

\(\displaystyle{y\frac{dy}{dx}=x^2+\sech^2(x)}\); \(y(0)=2\)  
solution 
Practice A11  

\(\displaystyle{\frac{dy}{dx}=\frac{x^2+1}{x^2(3y^2+1)}}\)  
solution 
Practice A12  

\(\displaystyle{\frac{dy}{dx}=\frac{42x}{3y^25}}\); \(y(1)=3\)  
solution 
Practice A13  

\(\displaystyle{\frac{dy}{dx}=2\sqrt{y}}\); \(y(0)=9\)  
solution 
Practice A14  

\(\displaystyle{\frac{dy}{dx}=\frac{x^2}{1y^2}}\)  
solution 
Practice A15  

\(\displaystyle{\frac{dy}{dx}=\frac{y\cos(x)}{1+2y^2}}\); \(y(0)=1\)  
solution 
Practice A16  

\(\displaystyle{\frac{du}{dt}=\frac{2t+\sec^2(t)}{2u}}\); \(u(0)=5\)  
solution 
Practice A17  

\(\displaystyle{ \frac{dy}{dx}=2x\sqrt{y} }\); \( y(0)=1 \)  
solution 
Practice A18  

\(\displaystyle{ \frac{dy}{dx}=e^{xy} }\); \( y(0)=\ln(2) \)  
solution 
Practice A19  

\(\displaystyle{\frac{dy}{dt} = \frac{t}{t^2y+y}}\), \(y(0)=3\), \(y >0\)  
answer 
solution 
Level B  Intermediate 
Practice B02  

\(\displaystyle{\frac{dy}{dx}=\frac{3x^2+4x+2}{2(y1)}}\); \(y(0)=1\)  
solution 