Reduction of Order for Linear Second-Order ODE
The Reduction of Order technique is a method for determining a second linearly independent solution to a homogeneous second-order linear ode given a first solution. This section has the following:
Example
It is best to describe the procedure with a concrete example. Consider the linear ode
We know that a solution to this problem is y1=exp(-3t). To obtain the general solution we need a second linearly independent solution to the problem.
We find the second solution by assuming
where v(t) is an unknown function. We now substitute this into the original ode (*) and derive a new ode for v(t). We have
and
Substituting these expressions into (*), we obtain
Many terms cancel in the above expression. We obtain
Either v''=0 or exp(-3t)=0. The second possibility only occurs if t=infinity and this is a degenerate case. We must have v''=0. In words: the second derivative of v is 0. This means that v is a line: v''=At+B, where A and B are constants. Hence, a second solution to the original ode (*) is
How do we choose A and B? Recall, our goal is determine a second linearly independent solution to the original ode (*). The first solution is y1=exp(-3t). Suppose we set A=0. Then y2=Bexp(-3t). In this case, y1 and y2 are multiples of each other, and are linearly dependent. On the other hand, suppose we choose B=0. Then y2=Atexp(-3t). In this case y1=exp(-3t) and y2=Atexp(-3t) are indeed linearly independent. As long as A is nonzero, y1 and y2 are linearly independent. It makes sense to set B=0, since if
the term Bexp-3t is the same as y1=exp(-3t) and adds nothing new.
To make a long story short(!), we choose A=1 and B=0. The second linearly independent solution to (*) is y2=texp(-3t) and the general solution to (*) is
General Solution Procedure
Let us now consider a general homogeneous linear ode:
and suppose f(t), which is known, is one solution of the ode. That means
We assume the second solution has the form:
where v(t) is unknown. We now substitute y2 into the original ode (**). We have
Substituting these expressions into (**), we have
Rearranging terms, we have
The term in parentheses is 0, since f(t) is a solution to (**). Hence, we are left with
Dividing by f, we have
This is a linear second-order ode where v'' depends on v' and t only and can be solved. It can be shown that
where
It follows that
Example
Consider the ode:
It can be shown that y1(t)=f(t)=t is a solution. We can apply formulas (****) and (***) to find the second solution. Here p(t)=-t. We have
Substituting this into (***), we have
This integral cannot be expressed in terms of elementary functions. The second linearly independent solution is
The general solution to the ode is
