5.6: Reduction of Order
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- Jul 20, 2020
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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′=u′y1+uy′1y″=u″y1+2u′y′1+uy″1
into Equation ??? yields
P0(x)(u″y1+2u′y′1+uy″1)+P1(x)(u′y1+uy′1)+P2(x)uy1=F(x).
Collecting the coefficients of u, u′, and u″ yields
(P0y1)u″+(2P0y′1+P1y1)u′+(P0y″1+P1y′1+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=2P0y′1+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
- 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.
- As a byproduct of (a), find a fundamental set of solutions of Equation ???.
Solution
a. If y=uex, then y′=u′ex+uex and y″=u″ex+2u′ex+uex, so
xy″−(2x+1)y′+(x+1)y=x(u″ex+2u′ex+uex)−(2x+1)(u′ex+uex)+(x+1)uex=(xu″−u′)ex.
Therefore y=uex is a solution of Equation ??? if and only if
(xu″−u′)ex=x2,
which is a first order equation in u′. We rewrite it as
u″−u′x=xe−x.
To focus on how we apply variation of parameters to this equation, we temporarily write z=u′, so that Equation ??? becomes
z′−zx=xe−x.
We leave it to you to show (by separation of variables) that z1=x is a solution of the complementary equation
z′−zx=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=vxwherev′x=xe−x,sov′=e−xandv=−e−x+C1.
Since u′=z=vx, u is a solution of Equation ??? if and only if
u′=vx=−xe−x+C1x.
Integrating this yields
u=(x+1)e−x+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=yp1−yp2=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
- Find the general solution of x2y″+xy′−y=x2+1, given that y1=x is a solution of the complementary equation x2y″+xy′−y=0. As a byproduct of this result, find a fundamental set of solutions of Equation ???.
- Solve the initial value problem x2y″+xy′−y=x2+1,y(1)=2,y′(1)=−3.
Solution
a. If y=ux, then y′=u′x+u and y″=u″x+2u′, so
x2y″+xy′−y=x2(u″x+2u′)+x(u′x+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=vx3wherev′x3=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=x3−1x−C12x2+C2.
Therefore the general solution of Equation ??? is
y=ux=x23−1−C12x+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=x23−1+c1x+c2x.
b. Differentiating Equation ??? yields
y′=2x3+c1−c2x2.
Setting x=1 in Equation ??? and Equation ??? and imposing the initial conditions y(1)=2 and y′(1)=−3 yieldsc1+c2=−83c1−c2=−113.
Solving these equations yields c1=−1/2, c2=19/6. Therefore the solution of Equation ??? isy=x23−1−x2+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
x2y″−3xy′+3y=0,
given that y1=x is a solution.
Solution
If y=ux then y′=u′x+u and y″=u″x+2u′, so
x2y″−3xy′+3y=x2(u″x+2u′)−3x(u′x+u)+3ux=x3u″−x2u′.
Therefore y=ux is a solution of Equation ??? if and only ifx3u″−x2u′=0.
Separating the variables u′ and x yieldsu″u′=1x,
soln|u′|=ln|x|+k,or equivalentlyu′=C1x.
Thereforeu=C12x2+C2,
so the general solution of Equation ??? isy=ux=C12x3+C2x,
which we rewrite asy=c1x+c2x3.
Therefore {x,x3} is a fundamental set of solutions of Equation ???.