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5.6: Reduction of Order

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In this section we give a method for finding the general solution of

P0(x)y+P1(x)y+P2(x)y=F(x)

if we know a nontrivial solution y1 of the complementary equation

P0(x)y+P1(x)y+P2(x)y=0.

The method is called reduction of order because it reduces the task of solving Equation ??? to solving a first order equation. Unlike the method of undetermined coefficients, it does not require P0, P1, and P2 to be constants, or F to be of any special form.

By now you shoudn’t be surprised that we look for solutions of Equation ??? in the form

y=uy1

where u is to be determined so that y satisfies Equation ???. Substituting Equation ??? and

y=uy1+uy1y=uy1+2uy1+uy1

into Equation ??? yields

P0(x)(uy1+2uy1+uy1)+P1(x)(uy1+uy1)+P2(x)uy1=F(x).

Collecting the coefficients of u, u, and u yields

(P0y1)u+(2P0y1+P1y1)u+(P0y1+P1y1+P2y1)u=F.

However, the coefficient of u is zero, since y1 satisfies Equation ???. Therefore Equation ??? reduces to

Q0(x)u+Q1(x)u=F,

with

Q0=P0y1andQ1=2P0y1+P1y1.

(It isn’t worthwhile to memorize the formulas for Q0 and Q1!) Since Equation ??? is a linear first order equation in u, we can solve it for u by variation of parameters as in Section 1.2, integrate the solution to obtain u, and then obtain y from Equation ???.

Example 5.6.1

  1. Find the general solution of xy(2x+1)y+(x+1)y=x2, given that y1=ex is a solution of the complementary equation xy(2x+1)y+(x+1)y=0.
  2. As a byproduct of (a), find a fundamental set of solutions of Equation ???.
Solution

a. If y=uex, then y=uex+uex and y=uex+2uex+uex, so

xy(2x+1)y+(x+1)y=x(uex+2uex+uex)(2x+1)(uex+uex)+(x+1)uex=(xuu)ex.

Therefore y=uex is a solution of Equation ??? if and only if

(xuu)ex=x2,

which is a first order equation in u. We rewrite it as

uux=xex.

To focus on how we apply variation of parameters to this equation, we temporarily write z=u, so that Equation ??? becomes

zzx=xex.

We leave it to you to show (by separation of variables) that z1=x is a solution of the complementary equation

zzx=0

for Equation ???. By applying variation of parameters as in Section 1.2, we can now see that every solution of Equation ??? is of the form

z=vxwherevx=xex,sov=exandv=ex+C1.

Since u=z=vx, u is a solution of Equation ??? if and only if

u=vx=xex+C1x.

Integrating this yields

u=(x+1)ex+C12x2+C2.

Therefore the general solution of Equation ??? is

y=uex=x+1+C12x2ex+C2ex.

b. By letting C1=C2=0 in Equation ???, we see that yp1=x+1 is a solution of Equation ???. By letting C1=2 and C2=0, we see that yp2=x+1+x2ex is also a solution of Equation ???. Since the difference of two solutions of Equation ??? is a solution of Equation ???, y2=yp1yp2=x2ex is a solution of Equation ???. Since y2/y1 is nonconstant and we already know that y1=ex is a solution of Equation ???, Theorem 5.1.6 implies that {ex,x2ex} is a fundamental set of solutions of Equation ???.

Although Equation ??? is a correct form for the general solution of Equation ???, it is silly to leave the arbitrary coefficient of x2ex as C1/2 where C1 is an arbitrary constant. Moreover, it is sensible to make the subscripts of the coefficients of y1=ex and y2=x2ex consistent with the subscripts of the functions themselves. Therefore we rewrite Equation ??? as

y=x+1+c1ex+c2x2ex

by simply renaming the arbitrary constants. We’ll also do this in the next two examples, and in the answers to the exercises.

Example 5.6.2

  1. Find the general solution of x2y+xyy=x2+1, given that y1=x is a solution of the complementary equation x2y+xyy=0. As a byproduct of this result, find a fundamental set of solutions of Equation ???.
  2. Solve the initial value problem x2y+xyy=x2+1,y(1)=2,y(1)=3.
Solution

a. If y=ux, then y=ux+u and y=ux+2u, so

x2y+xyy=x2(ux+2u)+x(ux+u)ux=x3u+3x2u.

Therefore y=ux is a solution of Equation ??? if and only if

x3u+3x2u=x2+1,

which is a first order equation in u. We rewrite it as

u+3xu=1x+1x3.

To focus on how we apply variation of parameters to this equation, we temporarily write z=u, so that Equation ??? becomes

z+3xz=1x+1x3.

We leave it to you to show by separation of variables that z1=1/x3 is a solution of the complementary equation

z+3xz=0

for Equation ???. By variation of parameters, every solution of Equation ??? is of the form

z=vx3wherevx3=1x+1x3,sov=x2+1andv=x33+x+C1.

Since u=z=v/x3, u is a solution of Equation ??? if and only if

u=vx3=13+1x2+C1x3.

Integrating this yields

u=x31xC12x2+C2.

Therefore the general solution of Equation ??? is

y=ux=x231C12x+C2x.

Reasoning as in the solution of Example 5.6.1a, we conclude that y1=x and y2=1/x form a fundamental set of solutions for Equation ???.

As we explained above, we rename the constants in Equation ??? and rewrite it as

y=x231+c1x+c2x.

b. Differentiating Equation ??? yields

y=2x3+c1c2x2.

Setting x=1 in Equation ??? and Equation ??? and imposing the initial conditions y(1)=2 and y(1)=3 yields

c1+c2=83c1c2=113.

Solving these equations yields c1=1/2, c2=19/6. Therefore the solution of Equation ??? is

y=x231x2+196x.

Using reduction of order to find the general solution of a homogeneous linear second order equation leads to a homogeneous linear first order equation in u that can be solved by separation of variables. The next example illustrates this.

Example 5.6.3

Find the general solution and a fundamental set of solutions of

x2y3xy+3y=0,

given that y1=x is a solution.

Solution

If y=ux then y=ux+u and y=ux+2u, so

x2y3xy+3y=x2(ux+2u)3x(ux+u)+3ux=x3ux2u.

Therefore y=ux is a solution of Equation ??? if and only if

x3ux2u=0.

Separating the variables u and x yields

uu=1x,

so

ln|u|=ln|x|+k,or equivalentlyu=C1x.

Therefore

u=C12x2+C2,

so the general solution of Equation ??? is

y=ux=C12x3+C2x,

which we rewrite as

y=c1x+c2x3.

Therefore {x,x3} is a fundamental set of solutions of Equation ???.


This page titled 5.6: Reduction of Order is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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