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.

Proof

Since L is a linear transformation, 

        L(y_h + y_p)  =  c₁L(y₁) + c₂L(y₂) + L(y_h) 

        =  c₁(0) + c₂(0) + g(t)  =  g(t)

This establishes that y_h + y_p 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  =  y_h + y_p 

for some constants c₁ and c₂ with

        y_h  =  c₁y₁ + c₂y₂  

This is equivalent to 

        y₃ - y_p  =  y_h 

We have 

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

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

        y₃ - y_p  =  c₁y₁ + c₂y₂  =  y_h 

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

Step 1:  Find the general solution y_h to the homogeneous differential equation.

Step 2:  Find a particular solution y_p to the nonhomogeneous differential equation.

Step 3:  Add y_h + y_p 

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.

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

        y_h  =  c₁e^-2t + c₂e^t 

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

        y_p  =  A e^-t sin t + B e^-t cos t

and find

        y_p'  =  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

        y_p''  =  -(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 

        y_p  =  -3/10 e^-t sin t + 1/10 e^-t cos t

Adding the particular solution to the homogeneous solution gives

        y  =  y_h + y_p  =  c₁e^-2t + c₂e^t + -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 

        y_h  =  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 

        y_p  =  At sin t  + Bt cos t

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

        y_p''  =  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 

        y_p  =  -2/5 cos t

Adding y_p to y_h gives

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

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