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Mathematics LibreTexts

3.4: Undetermined Coefficients

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

\[ L(y_1)  =  L(y_2)  =  0 \;\;\;  \text{and} \;\;\;  L(y_p)  =  g(t). \]

and let

\[   y_h  =  c_1y_1 + c_2y_2 \]

Then the general solution is given by 

\[  y  =  y_h +  y_p \].

Proof

Since \(L\) is a linear transformation,

\[ L(y_h + y_p) = C_1L(y_1) + C_2L(y_2) + L(y_h)\]

\[ = C_1(0) + C_2(0) + g(t) = g(t) \]

This establishes that \(y_h + y_p\)is a solution. Next we need to show that all solutions are of this form. Suppose that \(y_3\) is a solution to the nonhomogeneous differential equation. Then we need to show that 

\[ y_3 = y_h + y_p \]

for some constants \(c_1\) and \(c_2\) with

\[ y_h = c_1y_1 + c_2y_2 \]

This is equivalent to 

\[ y_3 - y_p = y_h \]

We have

\[ L(y_3 - y_p) = L(y_3) - L(y_p) = g(t) - g(t) = 0 \] 

Therefore \(y_3 - y_p\) is a solution to the homogeneous solution.  We can conclude that

\[y_3 - y_p = c_1y_1 + c_2y_2 = y_h\] 

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 \( g(t) \) 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 1

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 \( \left \{ \sin(3t), \cos(3t), t \sin(3t), t \cos(3t) \right \} \) 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 

\[ {f_1(t), f_2(t), ...f_n(t)} \]

Then there exists a whole number s such that 

\[  y_p  =  t^s[c_1f_1(t) + c_2f_2(t) + ... + c_nf_n(t)] \]

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 2

Find the general solution of the differential equation

\[  y'' + y' - 2y   =  e^{-t} \text{sin}\, t \]

Solution

First find the solution to the homogeneous differential equation

\[ y'' + y' - 2y = 0 \]

We have

\[ r^2 + r - 2 = (r - 1)(r + 2) = 0 \]

\[ r = -2 \;\;\; \text{or} \;\;\; r = 1\]

Thus

\[ y_h = c_1 e^{-2t} + c_2 e^t \]

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

\[\begin{align} y''_p &= -(A + B)(-e^{-t} \sin t + e^{-t} \cos t ) + (A - B)(-e^{-t} \cos t - e^{-t} \sin t ) \\  &= [(A + B) - (A - B)] e^{-t} \sin t + [-(A + B) - (A - B) ] e^{-t} \cos t \\ &= 2B e^{-t} \sin t - 2A e^{-t} \cos t  \end{align}\]

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

\[-3A + B = 1 \;\;\; \text{and} \;\;\; -A - 3B = 0\]

This system has solution 

\[ A = - \frac {3}{10}, \;\;\;    B = \frac{1}{10}. \]

The particular solution is 

\[ y_p = - \frac {3}{10} e^{-t} \sin t + \frac {1}{10} e^{-t} \cos t \]

Adding the particular solution to the homogeneous solution gives

\[ y = y_h + y_p = c_1 e^{-2t} + c_2 e^{t} + - \frac {3}{10} e^{-t} \sin t + \frac {1}{10} e^{-t} \cos t \]

Example 3

Solve

\[ y'' + y  =  5 \, \sin t\]

Solution

The characteristic equation is 

\[ r^2 + 1  =  0 \]

Which has the complex roots

\[ r = i \;\;\; \text{or} \;\;\;  r = -i \]

The homogeneous solution is 

\[        y_h  =  c_1 \text{sin} t + c_2 \text{cos} t \]

The UC-Set for \(\sin t\) is 

    \( \left \{ \sin t , \cos t \right \} \)    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

\[ \left \{ t \sin t, t \cos t \right \} \]

The particular solution is 

\[ y_p  =  At \, \sin t  + B \cos t \]

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

\[ y_p''  =  A \cos t + A \cos t - At \sin t - B\,  sin t - B\sin t - Bt \cos t  =  2A \cos t - At \sin t - 2B \sin t - Bt \cos t \]

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

\[ 2A = 0 \;\;\; \text{and} \;\;\;      -2 B = 5. \]

So

\[ A = 0  \;\;\;  \text{and} \;\;\;  B = - \frac {2}{5}. \]

We have

\[ y_p = - \frac {2}{5} \cos t. \]

Adding \( y_p\) to \(y_h \) gives

\[ y = c_1 \sin t + c_2 \cos t - \frac {2}{5} \cos t. \]

 

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