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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.

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

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 first-order, 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.

On the second-order, 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.)

• Example
• Solution