**Azərbaycan Dövlət Neft və Sənaye Universiteti**
**Differential Equations**
**Reduction of Order**
**In the preceding lecture we saw that the general solution of a homogeneous linear second-order differential equation**
**is a linear combination**
**where and are solutions that constitute a linearly independent set on some interval .**
**Beginning in the next lecture we examine a method for determining these solutions when the coefficients of the differential equation in are constants.**
**This method yields only a single solution of the DE. It turns out that we can construct a second solution of a homogeneous equation provided that we know a nontrivial solution of the DE. **
**The basic idea described in this lecture is that equation can be reduced to a linear first-order DE by means of a substitution involving the known solution .**
**A second solution of (1) is apparent after this first-order differential equation is solved.**
**Suppose that denotes a nontrivial solution of (1) and that is defined on an interval .**
**We seek a second solution so that the set consisting of and is linearly independent on .**
**Recall that if and are linearly independent, then their quotient is nonconstant on — that is**
** or **
**The function can be found by substituting into the given differential equation.**
**This method is called reduction of order because we must solve a linear first-order differential equation to find .**
**Example 1. A Second Solution by Reduction of Order**
**Given that is a solution of**
**on the interval use reduction of order to find a second solution .**
**Solution**
**If then the Product Rule gives**
**And so**
**Since the last equation requires**
**Solution (continued)**
**If we make the substitution this linear second-order equation in becomes**
**which is a linear first-order equation in .**
**Using the integrating factor we can write .**
**Integrating again then yields**
**Thus**
**Solution (continued)**
**After integrating we get or .**
**Solution (continued)**
**By picking and we obtain the desired second solution**
**Because for every the solutions are linearly independent on .**
**Since we have shown that and are linearly independent solutions of a linear second-order equation the expression in is actually the general solution of on .**
**Reduction of Order:**
**General Case**
**Suppose we divide by to put equation in the standard form**
**where and are continuous on some interval .**
**Let us suppose further that is a known solution of on and that for every in the interval.**
**If we define it follows that**
**And so**
**This implies that we must have**
**or **
**where we have let .**
**Observe that the last equation in is both linear and separable.**
**Separating variables and integrating we obtain**
**or**
**We solve the last equation for use and integrate again.**
**We thus obtain**
**By choosing and we find from that a second solution of equation is**
**Homework**
**It makes a good review of differentiation to verify that the function defined in satisfies equation and that are linearly independent on any interval on which is not zero.**
**Example 2. A Second Solution by Formula **
**The function is a solution of**
**Find the general solution of the differential equation on the interval .**
**Solution**
**Build the standard form of the differential equation**
**Solution**
**We then find from **
**The general solution on the interval is given by**
**Remark**
**Reduction of order can be used to find the general solution of a nonhomogeneous equation**
**whenever a solution of the associated homogeneous equation is known.**
**Thank you for attention**
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