This technique takes a secondorder, linear differential equation with one known solution and reduces the equation to a firstorder, linear equation that may possibly be solved using a firstorder technique. In general, the idea can be applied to even higher order equations but, due to complexity of integration, this technique is really only used on secondorder equations. However, it can still be useful in many cases.
classification: secondorder, linear, homogeneous 

\(y'' + p(t)y' + q(t)y = 0\) 
The idea is that we have an equation of the form \(y'' + p(t)y' + q(t)y = 0\) and we know one solution, \(y_1(t)\). To get a second solution, we try \(y_2=v(t)y_1(t)\), i.e. we assume that the second solution is equal to the first solution times some function of t. This may seem unusual but it works in many cases. Before we get into the details, let's watch this video.
video by Michel vanBiezen 

Now, let's work through the calculus and see where it gets us.
For simplicity, letting \(y=y_2\), we have \(y=v(t)y_1(t)\). Taking the derivative with respect to t, we get
\(y'=v'y_1+vy_1'\) and \(y''=v''y_1+2v'y_1'+vy_1''\).
To summarize, we have
\(y=vy_1\) 
\(y'=v'y_1+vy_1'\) 
\(y''=v''y_1+2v'y_1'+vy_1''\) 
Next, we take these expressions and substitute them back into the original differential equation \(y'' + p(t)y' + q(t)y = 0\). After collecting terms, we get
\( y_1v'' + (2y_1'+py_1)v' + (y_1''+py_1'+qy_1)v = 0 \)
Notice that the expression in parentheses of the last term is just the original differential equation, which is zero. So we end up with
\( y_1v'' + (2y_1'+py_1)v' = 0 \)
Now, let \(w=v'\) in the last equation to get \( y_1w' + (2y_1'+py_1)w = 0 \). This is now a firstorder, linear equation, which we can use to solve for w. Finally, we integrate w to get v and we are done.
Let's take a minute to watch the next video, which explains why this works.
video by PatrickJMT 

Specific Case  Constant Coefficients with Real, Repeated Roots
On the secondorder, linear page, we introduced the idea that, in the case of real, repeated roots, we would have one solution \(y_1=e^{rt}\). To show that the second solution is \(y_2=ty_1\), we will use reduction of order in this example. (Test yourself by trying this on your own before looking at the solution.)
Let \(y_1=c_1e^{rt}\) be one solution to \(ay''+by'+cy=0\). Since r is a repeated root of the characteristic equation, we can write the differential equation as \(y''2ry'+r^2y=0\). Use reduction of order to derive the second solution \(y_2=c_2te^{rt}\).
Let \(y_1=c_1e^{rt}\) be one solution to \(ay''+by'+cy=0\). Since r is a repeated root of the characteristic equation, we can write the differential equation as \(y''2ry'+r^2y=0\). Use reduction of order to derive the second solution \(y_2=c_2te^{rt}\).
Solution 

For simplicity, we will write \(y_2=v(t)y_1(t)=v(t)c_1e^{rt}\) as \(y=cve^{rt}\).
Our first step is to calculate the derivatives. We will need to use the product rule.
\(y=cve^{rt}\) 
\(y'=cvre^{rt}+cv'e^{rt} =\) \(ce^{rt}(vr+v')\) 
\(y''=ce^{rt}[v'r+v''] + cre^{rt}[vr+v'] = \) \(ce^{rt}[v'r+v''+vr^2+v'r] =\) \(ce^{rt}[v''+2rv'+r^2v]\) 
Now we substitute each of these into the original differential equation \(y''2ry'+r^2y=0\).
\(ce^{rt}[v''+2rv'+r^2v] \) \(2rce^{rt}[vr+v']+r^2cve^{rt}=0\)
\(ce^{rt}[v''+2rv'+r^2v\) \(2r^2v2rv'+r^2v]=0\)
The \(v\) and \(v'\) terms cancel, leaving only \(ce^{rt}v''=0\). Since \(ce^{rt}\) is never zero, we are left with \(v''=0\).
Now we integrate twice to get
\(v''=0 \to v'=k \to v=kt+p\) where k and p are the constants of integration.
Now we have \(y_2=ce^{rt}[kt+p]\). Next, we combine the two solutions into one. We were not asked to do this in the problem statement but doing this will allow us to explain what to do with all the constants.
\(
\begin{array}{rcl}
y &=& y_1+y_2 = ce^{rt}+ckte^{rt}+cpe^{rt} \\
&=& e^{rt}(c+cp)+kc(te^{rt}) \\
&=& c_1e^{rt}+c_2te^{rt} \\
&=& e^{rt}[c_1+c_2t]
\end{array}
\)
In the work above, we let \(c_1=c+cp\) and \(c_2=kc\), in essence just renaming the constants.
final answers  

second solution  \(y_2=c_2te^{rt}\) 
general solution  \(y=e^{rt}[c_1+c_2t]\) 
Practice
Unless otherwise instructed, find the second solution and the general solution to each differential equation using reduction of order. If initial conditions are given, solve the initial value problem also. Give your answers in exact terms and completely factored.
Solve \( x^2y'' + 5xy'  5y = 0, x > 0; y_1 = x \)
Problem Statement 

Solve \( x^2y'' + 5xy'  5y = 0, x > 0; y_1 = x \)
Solution 

video by PatrickJMT 

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Solve \( x^2y''  3xy' + 4y = 0; y_1 = x^2 \)
Problem Statement 

Solve \( x^2y''  3xy' + 4y = 0; y_1 = x^2 \)
Solution 

video by MIP4U 

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Solve \( t^2y''  t(t+2)y' + (t+2)y = 0; y_1 = t \)
Problem Statement 

Solve \( t^2y''  t(t+2)y' + (t+2)y = 0; y_1 = t \)
Solution 

Here are two videos solving this problem from two different instructors.
video by Michel vanBiezen 

video by Educator.com 

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Solve \( x^2y'' + 3xy' + y = 0; \) \( y_1 = 1/x \)
Problem Statement 

Solve \( x^2y'' + 3xy' + y = 0; \) \( y_1 = 1/x \)
Solution 

video by MIP4U 

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Solve \( 4x^2y'' + y = 0; y_1 = \sqrt{x} \)
Problem Statement 

Solve \( 4x^2y'' + y = 0; y_1 = \sqrt{x} \)
Solution 

This is a CauchyEuler equation. You can find more discussion on CauchyEuler equations on the this 17calculus page.
video by blackpenredpen 

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You CAN Ace Differential Equations
external links you may find helpful 

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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Single Variable Calculus 

MultiVariable Calculus 

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Practice Instructions
Unless otherwise instructed, find the second solution and the general solution to each differential equation using reduction of order. If initial conditions are given, solve the initial value problem also. Give your answers in exact terms and completely factored.