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Differential Equations: Method of Undetermined Coefficients

Problem 5

\[ y'' - 2y' - 3y = -3te^{-t} \]

Homogeneous solution:

\[ y_1 = e^{-t}, \quad y_2 = e^{3t} \]
\[ y = c_1 e^{-t} + c_2 e^{3t} + Y \]

Analyzing the Right-Hand Side (RHS)

RHS: \( te^{-t} \)

  • \( t \) is a first degree polynomial.
  • \( e^{-t} \) is an exponential term.

Trial Particular Solution:

\[ Y = t(At + B)e^{-t} = Ate^{-t} + Be^{-t} \]

Note on Duplication:

The term \( Be^{-t} \) is duplicating \( y_1 \). Therefore, we must multiply the trial solution by \( t \).

\[ Y = At^2 e^{-t} + Bte^{-t} \]
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Problem 21

\[ y'' + 3y' = 2t^4 + t^2 e^{-3t} + \sin 3t \]

Homogeneous solution:

\[ y_1 = 1, \quad y_2 = e^{-3t} \]
\[ y = c_1 + c_2 e^{-3t} + Y \]

Trial Particular Solution:

\[ Y = t(At^4 + Bt^3 + Ct^2 + Dt + E) + t(Ft^2 + Gt + H)e^{-3t} + I \sin 3t + J \cos 3t \]

Repeated Roots

Case 1:

\[ y'' - 4y = 0 \]\[ r^2 = 4 \implies r = \pm 2 \]

Case 2:

\[ y'' + 2y' + y = 0 \]\[ r^2 + 2r + 1 = 0 \]\[ r = -1, -1 \]

For repeated roots \( r = -1 \):

\[ y_1 = e^{-t} \]\[ y_2 = ve^{-t} \quad \text{where } v = At + B \]\[ y_2 = Ate^{-t} + Be^{-t} \]

General Solution:

\[ y = c_1 e^{-t} + c_2(Ate^{-t} + Be^{-t}) \]\[ y = C_1 e^{-t} + C_2 te^{-t} \]
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3.6 Variation of Parameters

Another way to solve nonhomogeneous eqs.

\[ y'' - 5y' + 6y = 2e^t \]

First, find homogeneous solutions: \( y_1 = e^{2t} \), \( y_2 = e^{3t} \)

Normally, \( y = c_1 y_1 + c_2 y_2 + Y \)

parameters: \( c_1, c_2 \)

But now we look at it as: \( y = u_1(t)y_1 + u_2(t)y_2 \)

Find \( u_1, u_2 \) by subbing into DE:

\[ y = u_1 e^{2t} + u_2 e^{3t} \]\[ y' = 2u_1 e^{2t} + u_1' e^{2t} + 3u_2 e^{3t} + u_2' e^{3t} \]

Looking ahead: Two unknowns \( (u_1, u_2) \) ONE eq. (DE) \( \rightarrow \) impose extra condition:

\[ u_1' e^{2t} + u_2' e^{3t} = 0 \]

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\[ y' = 2u_1 e^{2t} + 3u_2 e^{3t} \]\[ y'' = 4u_1 e^{2t} + 2u_1' e^{2t} + 9u_2 e^{3t} + 3u_2' e^{3t} \]

DE: \( y'' - 5y' + 6y = 2e^t \)

\[ 4u_1 e^{2t} + 2u_1' e^{2t} + 9u_2 e^{3t} + 3u_2' e^{3t} \]\[ - 10u_1 e^{2t} - 15u_2 e^{3t} \]\[ + 6u_1 e^{2t} + 6u_2 e^{3t} = 2e^t \]

Condition imposed earlier:

DE:

\[ \begin{cases} u_1' e^{2t} + u_2' e^{3t} = 0 & - \text{ ①} \\ 2u_1' e^{2t} + 3u_2' e^{3t} = 2e^t & - \text{ ②} \end{cases} \]

mult. ① by 2

\[ 2u_1' e^{2t} + 2u_2' e^{3t} = 0 \quad - \text{ ③} \]

② - ③

\[ u_2' e^{3t} = 2e^t \]
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\[ u_2' = 2e^{-2t} \]\[ u_2 = -e^{-2t} + C_2 \]

from (1)

\[ u_1' e^{2t} = -u_2' e^{3t} \]\[ = -2e^{-2t} e^{3t} = -2e^t \]\[ u_1 = -2e^t + C_1 \]

Solution:

\[ y = u_1 y_1 + u_2 y_2 \]\[ = (-2e^t + C_1) e^{2t} + (-e^{-2t} + C_2) e^{3t} \]\[ = C_1 e^{2t} + C_2 e^{3t} - 2e^{3t} - e^t \]
\[ y = \underbrace{C_1 e^{2t} + C_2 e^{3t}}_{\text{homogeneous}} - \underbrace{e^t}_{\text{particular}} \]
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for ANY \[ y'' + p(t)y' + q(t)y = g(t) \] we always end up solving

\[ u_1' y_1 + u_2' y_2 = 0 \]\[ u_1' y_1' + u_2' y_2' = g(t) \]

which is equivalent to the matrix form:

\[ \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \begin{bmatrix} u_1' \\ u_2' \end{bmatrix} = \begin{bmatrix} 0 \\ g(t) \end{bmatrix} \]

unique solution if

\[ \underbrace{\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \neq 0}_{\text{Wronskian}} \]

Example

\[ y'' - 2y' + y = e^t \]

homogeneous: \[ y_1 = e^t \quad y_2 = t e^t \]

if using undetermined coeff., \( Y \) has duplications

using variation of parameters, \( u_1, u_2 \) take care of duplications

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Solving the System of Equations

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\[ \begin{aligned} u_1' e^t + u_2' t e^t &= 0 & \text{--- (1)} \\ u_1' e^t + u_2' (t e^t + e^t) &= e^t & \text{--- (2)} \end{aligned} \]

Step 1: Subtract (1) from (2)

Performing the operation \( (2) - (1) \):

\[ \begin{aligned} u_2' e^t &= e^t \\ u_2' &= 1 \\ u_2 &= t + c_2 \end{aligned} \]

Step 2: Solve for \( u_1 \)

From equation (1):

\[ \begin{aligned} u_1' &= -u_2' t \\ &= -t \\ u_1 &= -\frac{1}{2} t^2 + c_1 \end{aligned} \]

General Solution

\[ \begin{aligned} y &= u_1 y_1 + u_2 y_2 \\ &= \left( -\frac{1}{2} t^2 + c_1 \right) e^t + (t + c_2) t e^t \\ &= c_1 e^t + c_2 t e^t - \frac{1}{2} t^2 e^t + t^2 e^t \\ &= \underbrace{c_1 e^t + c_2 t e^t}_{\text{homogeneous}} + \underbrace{\frac{1}{2} t^2 e^t}_{\text{particular}} \end{aligned} \]
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Example: Variation of Parameters

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\[ (t^2 - 1) y'' - 2t y' + 2y = (t^2 - 1)^2 \]

Given fundamental solutions: \( y_1 = t \) and \( y_2 = t^2 + 1 \).

Note: \( (t^2 - 1)^2 \) is NOT \( g(t) \) yet.

Standard Form

Divide by the leading coefficient to put the differential equation in standard form:

\[ y'' - \frac{2t}{t^2 - 1} y' + \frac{2}{t^2 - 1} y = \boxed{t^2 - 1} \]

\( g(t) = t^2 - 1 \)

RHS in DE in standard form where \( y'' \) has 1 as coefficient.

System for Variation of Parameters

\[ \begin{aligned} u_1' y_1 + u_2' y_2 &= 0 \\ u_1' y_1' + u_2' y_2' &= g(t) \end{aligned} \]