Undetermined Coefficients

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)  =  0          and            L(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 

        y3 - yp  =  yh 

We have 

        L(y3 - yp)  =  L(y3) - L(yp)  =  g(t) - g(t)  =  0

Therefore y3 - yp is a solution to the homogeneous solution.  We can conclude that 

        y3 - yp  =  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.

 

Etample

Let g(t)  =  t sin(3t)

Then

        g'(t)  =  sin(3t) + 3t cos(3t)                g''(t)  =  6 cos(3t) - 9t sin(3t)

        g(3)(t)  =  -27 sin(3t) - 27t cos(3t)       g(4)(t)  =  81 cos(3t) - 108t sin(3t)       

        g(4)(t)  =  405 sin(3t) + 243t cos(3t)       g(5)(t)  =  1458 cos(3t) - 729t cos(3t)       

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

        g(n)(t)  =  A sin(3t) + B cos(3t) + Ct sin(3t) + Dt cos(3t)

We can say that {sin(3t), cos(3t), t sin(3t), t cos(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

Find the general solution of the differential equation

        y'' + y' - 2y   =  e-t sin t

 

Solution

First find the solution to the homogeneous differential equation

        y'' + y' - 2y  =  0

We have

        r2 + r - 2  =  (r - 1)(r + 2)    =  0

        r  =  -2        or         r  =  1

Thus

        yh  =  c1e-2t + c2et 

Next notice that e-t sin t and all of its derivatives are of the form 

        A e-t sin t + B e-t cos t

We set

        yp  =  A e-t sin t + B e-t cos t

and find

        yp'  =  A (-e-t sin t + e-t cos t) + B (-e-t cos t - e-t sin t)  

        =  -(A + B)e-t sin t + (A - B)e-t cos t

and

        yp''  =  -(A + B)(-e-t sin t + e-t cos t) + (A - B)(-e-t cos t - e-t sin t)

        =  [(A + B) - (A - B)]e-t sin t  +  [-(A + B) - (A - B)]e-t cos t

        =  2B e-t sin t - 2A e-t cos t

Now put these into the original differential equation to get

        2B e-t sin t - 2A e-t cos t + -(A + B)e-t sin t + (A - B)e-t cos t - 2(A e-t sin t + B e-t cos t)   =  e-t sin t

Combine like terms to get

        (2B - A - B - 2A) e-t sin t + (-2A + A - B - 2B) e-t cos t  =  e-t sin t

or

        (-3A + B) e-t sin t + (-A - 3B) e-t cos t  =  e-t sin t

Equating coefficients, we get

        -3A + B  =  1

        -A - 3B  =  0

This system has solution 

        A  =  -3/10        B  =  1/10

The particular solution is 

        yp  =  -3/10 e-t sin t + 1/10 e-t cos t

Adding the particular solution to the homogeneous solution gives

        y  =  yh + yp  =  c1e-2t + c2et + -3/10 e-t sin t + 1/10 e-t cos t

 

Example

Solve

        y'' + y  =  5 sin t

Solution

The characteristic equation is 

        r2 + 1  =  0

Which has the complex roots

        r  =  i        or         r  =  -i

The homogeneous solution is 

        yh  =  c1sin t + c2 cos t

The UC-Set for sin t is 

        {sin t, 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

        {t sin t, t cost}

The particular solution is 

        yp  =  At sin t  + Bt cos t

        yp'  =  A sin t + At cos t + B cos t - Bt sin t

        yp''  =  A cos t + A cos t - At sin t - B sin t - B sin t - Bt cos t  =  2A cos t - At sin t - 2B sin t - Bt cos t

Now put these back into the original differential equation to get

        2A cos t - At sin t - 2B sin t - Bt cos t + At sin t  + Bt cos t  =  5 sin t

        2A cos t  - 2B sin t  =  5 sin t

Equating coefficients gives

        2A  =  0        -2B  =  5

So

        A  =  0  and  B  =  -2/5

We have 

        yp  =  -2/5 cos t

Adding yp to yh gives

        y  =  c1sin t + c2 cos t - 2/5 cos t

 


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