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Reduction of Order 

classification: secondorder, linear, homogeneous 
\(y'' + p(t)y' + q(t)y = 0\) 
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.
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. 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 a video that goes through these equations again.
PatrickJMT  Reduction of Order  Why It Works  
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.
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Practice Problems 

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.