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3.4: Method of Undetermined Coefficients

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Up to now, we have considered homogeneous second order differential equations. In this discussion, we will investigate nonhomogeneous second order linear differential equations.

Theorem: Solutions of Nonhomogeneous Second Order Linear Differential Equations

Let

L(y)=y+p(t)y+q(t)y=g(t)

be a second order linear differential equation with p, q, and g continuous and let

L(y1)=L(y2)=0andL(yp)=g(t)

and let

yh=c1y1+c2y2.

Then the general solution is given by

y=yh+yp.

Proof

Since L is a linear transformation,

L(yh+yp)=C1L(y1)+C2L(y2)+L(yh)=C1(0)+C2(0)+g(t)=g(t).

This establishes that yh+yp is a solution. Next we need to show that all solutions are of this form. Suppose that y3 is a solution to the nonhomogeneous differential equation. Then we need to show that

y3=yh+yp

for some constants c1 and c2 with

yh=c1y1+c2y2.

This is equivalent to

y3yp=yh.

We have

L(y3yp)=L(y3)L(yp)=g(t)g(t)=0.

Therefore y3yp is a solution to the homogeneous solution. We can conclude that

y3yp=c1y1+c2y2=yh.

This theorem provides us with a practical way of finding the general solution to a nonhomogeneous differential equation.

  • Step 1: Find the general solution yh to the homogeneous differential equation.
  • Step 2: Find a particular solution yp to the nonhomogeneous differential equation.
  • Step 3: Add yh+yp.

We have already learned how to do Step 1 for constant coefficients. We will now embark on a discussion of Step 2 for some special functions g(t).

Definition

A function g(t) generates a UC-set if the vector space of functions generated by g(t) and all the derivatives of g(t) is finite dimensional.

Example 3.4.1

Let g(t)=tsin(3t).

Then

g(t)=sin(3t)+3tcos(3t)g(t)=6cos(3t)9tsin(3t)g(3)(t)=27sin(3t)27tcos(3t)g(4)(t)=81cos(3t)108tsin(3t)g(4)(t)=405sin(3t)243tcos(3t)g(5)(t)=1458cos(3t)729tcos(3t)

We can see that g(t) and all of its derivative can be written in the form

g(n)(t)=Asin(3t)+Bcos(3t)+Ctsin(3t)+Dtcos(3t).

We can say that {sin(3t),cos(3t),tsin(3t),tcos(3t)} is a basis for the UC-Set.

We now state without proof the following theorem tells us how to find the particular solution of a nonhomogeneous second order linear differential equation.

Theorem:

Let

L(y)=ay+by+cy=g(t)

be a nonhomogeneous linear second order differential equation with constant coefficients such that g(t) generates a UC-Set

f1(t),f2(t),...fn(t)

Then there exists a whole number s such that

yp=ts[c1f1(t)+c2f2(t)+...+cnfn(t)]

is a particular solution of the differential equation.

Remark: The "s" will come into play when the homogeneous solution is also in the UC-Set.

Example 3.4.2

Find the general solution of the differential equation

y+y2y=etsint.

Solution

First find the solution to the homogeneous differential equation

y+y2y=0.

We have

r2+r2=(r1)(r+2)=0

r=2orr=1.

Thus

yh=c1e2t+c2et.

Next notice that etsint and all of its derivatives are of the form

Aetsint+Betcost.

We set

yp=Aetsint+Betcost

and find

yp=A(etsint+etcost)+B(etcostetsint)=(A+B)etsint+(AB)etcost

and

yp=(A+B)(etsint+etcost)+(AB)(etcostetsint)=[(A+B)(AB)]etsint+[(A+B)(AB)]etcost=2Betsint2Aetcost.

Now put these into the original differential equation to get

2Betsint2Aetcost+(A+B)etsint+(AB)etcost2(Aetsint+Betcost)=etsint.

Combine like terms to get

(2BAB2A)etsint+(2A+AB2B)etcost=etsint

or

(3A+B)etsint+(A3B)etcost=etsint.

Equating coefficients, we get

3A+B=1andA3B=0.

This system has solution

A=310,B=110.

The particular solution is

yp=310etsint+110etcost.

Adding the particular solution to the homogeneous solution gives

y=yh+yp=c1e2t+c2et+310etsint+110etcost.

Example 3.4.3

Solve

y+y=5sint.

Solution

The characteristic equation is

r2+1=0.

Which has the complex roots

r=iorr=i.

The homogeneous solution is

yh=c1sint+c2cost.

The UC-Set for sint is {sint,cost}. Derivatives are all sin and cos functions

Notice that both of the functions in the UC-Set are solutions to the homogeneous differential equation. We need to multiply by t to get

{tsint,tcost}.

The particular solution is

yp=Atsint+Bcostyp=Asint+Atcost+BcostBtsintyp=Acost+AcostAtsintBsintBsintBtcost=2AcostAtsint2BsintBtcost.

Now put these back into the original differential equation (Equation ???) to get

2AcostAtsint2BsintBtcost+Atsint+Btcost=5sint2Acost2Bsint=5sint.

Equating coefficients gives

2A=0and2B=5.

So

A=0andB=25.

We have

yp=25cost.

Adding yp to yh gives

y=c1sint+c2cost25cost.

Contributors and Attributions


This page titled 3.4: Method of Undetermined Coefficients is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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