# 3.4: Method of 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

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

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,

\[\begin{align*} L(y_h + y_p) &= C_1L(y_1) + C_2L(y_2) + L(y_h)\\[5pt] &= C_1(0) + C_2(0) + g(t) = g(t). \end{align*}\]

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

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

\(\square\)

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.

Example \(\PageIndex{1}\)

Let \( g(t) = t \sin(3t) \).

Then

\[\begin{align*} g'(t) &= \sin(3t) + 3t \cos(3t) & g''(t) &= 6 \cos(3t) - 9t \sin(3t) \\ g^{(3)} (t) &= -27 \sin(3t) - 27t \cos(3t) & g^{(4)}(t) &= 81 \cos(3t) - 108t \sin(3t) \\ g^{(4)} (t) &= 405 \sin(3t) - 243t \cos(3t) & g^{(5)}(t) &= 1458 \cos(3t) - 729t \cos(3t) \end{align*}\]

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 \(\PageIndex{2}\)

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

\[ \begin{align*} y'_p &= A ( -e^{-t} \sin t + e^{-t} \cos t) + B (-e^{-t} \cos t - e^{-t} \sin t ) \\[5pt] &= -(A + B)e^{-t} \sin t + (A - B)e^{-t} \cos t \end{align*}\]

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 \(\PageIndex{3}\)

Solve

\[ y'' + y = 5 \, \sin t. \label{ex3.1}\]

**Solution**

The characteristic equation is

\[ r^2 + 1 = 0. \nonumber\]

Which has the complex roots

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

The homogeneous solution is

\[ y_h = c_1 \sin t + c_2 \cos t. \nonumber \]

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

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

The particular solution is

\[\begin{align*} y_p &= At \, \sin t + B \cos t \\[5pt] y_p' &= A \sin t + At \cos t + B \cos t - Bt \sin t \\[5pt] y_p'' &= A \cos t + A \cos t - At \sin t - B\, \sin t - B\sin t - Bt \cos t \\[5pt]&= 2A \cos t - At \sin t - 2B \sin t - Bt \cos t. \end{align*}\]

Now put these back into the original differential equation (Equation \ref{ex3.1}) to get

\[\begin{align*} 2A \cos t - At \sin t -2B \sin t - Bt \cos t + At \sin t + Bt \cos t &= 5 \sin t \\[5pt] 2A \cos t - 2B \sin t &= 5 \sin t. \end{align*}\]

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

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

### Contributors

- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.