Sometimes we can solve a differential equation by using substitution and changing the variables. This changes the equation into one where we can use a technique we learned previously. There are several different kinds of substitutions that can be done. Let's get started with the easiest where the substitution is given in the problem statement.
Alternate Name For Substitution  Change of Variables 

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Given in the Problem Statement
This, of course, is the easiest kind of substitution you can have. You don't have to analyze or classify the differential equation and then make a determination of what type of substitution will help transform the equation into one that you can solve. The thing to remember here is to make sure and take the derivative of the substitution and substitute for the differentials based on what you are given. For example, if your original variable is x and the new one is t, you can't just say that \(dx = dt\). You need to calculate \(dt\) based on your equation. (You probably remember this from integration by substitution. The same rule applies here.)
Here is a video showing this technique with an example and lots of detail. This is great place to start to understand what is involved with the technique of substitution.
video by Dr Chris Tisdell 

Okay, after watching that video, you should have a clue what substitution is about and how to accomplish it. Although this technique is not complicated, let's work some practice problems, before we go on.
Practice
Unless otherwise instructed, solve these differential equations using the indicated substitution. Give your answers in exact form.
\(\displaystyle{ x\frac{dy}{dx} + y = xy^2 }\); \( y = 1/v \)
Problem Statement 

\(\displaystyle{ x\frac{dy}{dx} + y = xy^2 }\); \( y = 1/v \)
Solution 

video by Newcastle University Math 

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\( y' = x + y \); \( u = x + y \)
Problem Statement 

\( y' = x + y \); \( u = x + y \)
Solution 

video by Krista King Math 

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The trick now is to learn how to choose a substitution on your own, based on what the differential equation looks like.
Basic \(v=y'\)
In the special case when you have a second order equation with the dependent variable missing, you can use the substitution \(v=y'\) to reduce it to a first order equation, find \(v\) and then integrate to get \(y\). This technique can be used to solve these practice problems.
Practice
Unless otherwise instructed, solve these differential equations using the substitution \(v=y'\). Give your answers in exact form.
\( y'' + y' = t \)
Problem Statement 

\( y'' + y' = t \)
Final Answer 

\( y = t^2/2  t  c_1e^{t} + c_2 \)
Problem Statement 

\( y'' + y' = t \)
Solution 

Let \(w=y'\) and so \(w'=y''\).
\(y''+y'=t \to w'+w=t\)
This equation is now firstorder, linear. So we can use an integrating factor.
\(\mu=exp \int{1~dt} = e^t \)
\(\displaystyle{ \frac{d}{dt}[we^t] = te^t }\)
\(we^t = \int{ te^t~dt}\)
Now we use integration by parts on the integral.
\(u=t \to du=dt\) and \(dv=e^tdt \to v=e^t\)
\(\begin{array}{rcl}
we^t &=& te^t\int{e^t~dt} \\
&=& te^te^t+c_1 \\
w &=& t1+c_1e^{t} \\
y' &=& t1+c_1e^{t} \\
y &=& \int{t1+c_1e^{t}~dt} \\
y &=& t^2/2tc_1e^{t}+c_2
\end{array}\)
Final Answer 

\( y = t^2/2  t  c_1e^{t} + c_2 \)
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\( ty''  2y' = 12t^2 \); \(t > 0\)
Problem Statement 

\( ty''  2y' = 12t^2 \); \(t > 0\)
Final Answer 

\( y = 4t^3\ln t + kt^3 + c \)
Problem Statement 

\( ty''  2y' = 12t^2 \); \(t > 0\)
Solution 

Let \(w=y'\), so \(w'=y''\).
\(ty''2y'=12t^2 \to tw'2w=12t^2 \to \) \(w'(2/t)w=12t\)
This last equation is firstorder, linear, so we can use an integrating factor.
\(\mu=exp \int{2/t~dt} = e^{2\ln t}=t^{2}\)
\(\displaystyle{ \frac{d}{dt}[t^{2}w] = 12/t }\)
\( t^{2}w = 12\ln t + c_1 \) 
\( w = 12t^{2}\ln t+c_1t^2 \) 
\( y = \int{ 12t^2 \ln t+c_1t^2~dt} \) 
\(\displaystyle{ y = 12\left[ \frac{t^3}{3}\ln t  \int{t^2 / 3 ~dt} \right] + c_1(t^3)/3 }\) 
\( y = 4t^3\ln t  4t^3/3 +c_1t^3/3+c_2 \) 
\( y = 4t^3\ln t + kt^3+c \) 
Notes:
1. We used integration by parts on the integral \(\int{t^2\ln t~dt}\) letting
\(u=\ln t \to du=(1/t)dt\) and \(dv=t^2~dt \to v=t^3/3\).
2. In the last step, we combined the \(t^3\) terms and replaced the constant. This was not necessary but it made the final answer more compact and easier to read. Doing this step would also allow us to more easily determine the particular solution, if initial conditions had been given.
Final Answer 

\( y = 4t^3\ln t + kt^3 + c \)
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Scaling
Sometimes we may just want to scale the equation to make it easier to work with or make it dimensionless. In order to do that, we use these substitutions.
\(x_1 = x/a\)  \(y_1 = y/b\) 
where \(x\) and \(y\) are the old variables  
\(x_1\) and \(y_1\) are the new variables, and  
\(a\) and \(b\) are constants 
Let's watch a video clip, discussing this in detail.
video by MIT OCW 

Homogeneous ODEs
These first order, linear differential equations can be written in the form, \(y' = f(y/x)\), which should make it obvious that the substitution we use is \(z=y/x\). This is the most common form of substitution taught in first year differential equations. Let's watch a video clip discussing this.
video by MIT OCW 

Now try solving these homogeneous ODEs.
Practice
\(\displaystyle{ \frac{dy}{dx} = \frac{x^2+y^2}{2x^2} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} = \frac{x^2+y^2}{2x^2} }\)
Solution 

video by Newcastle University Math 

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\(\displaystyle{ \frac{dy}{dx} = \frac{y^2x^2}{xy} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} = \frac{y^2x^2}{xy} }\)
Solution 

video by PatrickJMT 

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\(\displaystyle{ \frac{dy}{dx} = \frac{y^3+y^2x+yx^2}{yx^2} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} = \frac{y^3+y^2x+yx^2}{yx^2} }\)
Solution 

video by PatrickJMT 

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\(\displaystyle{ \frac{dy}{dx} = \frac{y^4+yx^3}{x^4} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} = \frac{y^4+yx^3}{x^4} }\)
Solution 

video by PatrickJMT 

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\(\displaystyle{ \frac{dy}{dx} = \frac{y^3+2y^2xyx^2}{yx^2+x^3} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} = \frac{y^3+2y^2xyx^2}{yx^2+x^3} }\)
Solution 

video by PatrickJMT 

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\(\displaystyle{ \frac{dy}{dx} = \frac{x+y}{x} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} = \frac{x+y}{x} }\)
Solution 

video by Khan Academy 

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\(\displaystyle{ \frac{dy}{dx} = \frac{x^2+3y^2}{2xy} }\)
Problem Statement 

\(\displaystyle{ \frac{dy}{dx} = \frac{x^2+3y^2}{2xy} }\)
Solution 

video by Khan Academy 

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CauchyEuler Equation
When you have an equation of the form \(\displaystyle{ x^2\frac{d^2y}{dx^2} + ax\frac{dy}{dx} + by = 0, ~~~ x > 0 }\), where a and b are constants, you have an CauchyEuler equation. One substitution that works here is to let \(t = \ln(x)\). This substitution changes the differential equation into a second order equation with constant coefficients. We discuss this in more detail on a separate page.
As you can tell from the discussion above, there are many types of substitution problems, each with its own technique. We have touched on only a few here. Go through your textbook or search the internet and see if you can find others. If you need a good differential equations textbook, you can find several options at the 17calculus bookstore.
You CAN Ace Differential Equations
related topics on other pages 
external links you may find helpful 

Pauls Online Math Notes  Differential Equations Substitution 
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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