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 \(y_2=c_2te^{rt}\).
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  \(y=e^{rt}[c_1+c_2t]\) 
close solution 
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.
Problem Statement 

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

video by PatrickJMT 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by MIP4U 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

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

video by Educator.com 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

video by MIP4U 

close solution

Log in to rate this practice problem. 

Problem Statement 

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

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

close solution

Log in to rate this practice problem. 

You CAN Ace Differential Equations
external links you may find helpful 

To bookmark this page and practice problems, log in to your account or set up a free account.
Single Variable Calculus 

MultiVariable Calculus 

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
Help Keep 17Calculus Free 

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. 