Use The Annihilator Method To Find The General Solution Of The Differential Equation Y" - 2y 3y = E-!
When faced with a second-order linear differential equation, especially one involving nonhomogeneous terms like exponential functions, the Annihilator Method offers a powerful approach to find the general solution. This method combines the techniques of solving the homogeneous part of the differential equation and applying an appropriate annihilator to manage the nonhomogeneous term effectively. In this comprehensive guide, we'll explore how to utilize the Annihilator Method to solve the differential equation:
\[ Y'' - 2Y' + 3Y = e^{-x} \]
by breaking down each step, understanding the underlying principles, and providing practical examples.
Understanding the Structure of the Differential Equation
Before delving into the Annihilator Method, it’s essential to comprehend the components of the differential equation:
- Type: Second-order linear differential equation
- Homogeneous part: \( Y'' - 2Y' + 3Y = 0 \)
- Nonhomogeneous part: \( e^{-x} \)
The goal is to find the general solution \( Y(x) \), which combines:
- The complementary (homogeneous) solution \( Y_c \)
- The particular solution \( Y_p \)
The Annihilator Method simplifies finding \( Y_p \) by transforming the nonhomogeneous term into a solution of a differential operator, enabling us to solve the entire equation systematically.
Step 1: Find the Homogeneous Solution
Begin by solving the homogeneous differential equation:
\[ Y'' - 2Y' + 3Y = 0 \]
Characteristic Equation:
\[
r^2 - 2r + 3 = 0
\]
Solve for \( r \):
\[
r = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times 3}}{2} = \frac{2 \pm \sqrt{4 - 12}}{2} = \frac{2 \pm \sqrt{-8}}{2}
\]
\[
r = 1 \pm i \sqrt{2}
\]
Homogeneous Solution:
\[
Yc = e^{x} \left( C1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x) \right)
\]
where \( C1 \) and \( C2 \) are arbitrary constants.
Summary:
- The homogeneous solution accounts for the complementary part of the general solution.
- These solutions typically involve exponential and trigonometric functions when roots are complex.
---
Step 2: Identify the Nonhomogeneous Term and Its Nature
The nonhomogeneous term is:
\[ e^{-x} \]
This exponential function suggests that the particular solution should include an exponential term, possibly multiplied by polynomial functions if needed.
---
Step 3: Find the Appropriate Annihilator
The core idea of the Annihilator Method is to find a differential operator that, when applied to the nonhomogeneous term, results in zero. This operator is called the annihilator.
Identify the annihilator for \( e^{-x} \):
- The exponential function \( e^{ax} \) is annihilated by the differential operator \( D - a \), where \( D = \frac{d}{dx} \).
- For \( e^{-x} \), the annihilator is:
\[
D + 1
\]
because:
\[
(D + 1) e^{-x} = \frac{d}{dx} e^{-x} + e^{-x} = - e^{-x} + e^{-x} = 0
\]
Construct the combined differential operator:
- To transform the original differential equation into an equation with zero on the right side, apply the annihilator:
\[
(D + 1)(Y'' - 2Y' + 3Y) = 0
\]
- This produces a higher-order homogeneous differential equation.
---
Step 4: Formulate the Annihilated Differential Equation
Applying the operator:
\[
(D + 1)(Y'' - 2Y' + 3Y) = 0
\]
gives:
\[
(D + 1)(Y'') - 2(D + 1)Y' + 3(D + 1)Y = 0
\]
which expands to:
\[
Y''' + Y'' - 2Y' - 2Y + 3Y' + 3Y = 0
\]
Simplify:
\[
Y''' + Y'' + ( - 2Y' + 3Y' ) + ( - 2Y + 3Y ) = 0
\]
\[
Y''' + Y'' + Y' + Y = 0
\]
This is the annihilated differential equation.
Note:
- Since this is a third-order homogeneous differential equation, its general solution will include three arbitrary constants.
---
Step 5: Solve the Homogeneous Equation Derived from the Annihilator
Solve:
\[
Y''' + Y'' + Y' + Y = 0
\]
Characteristic Equation:
\[
r^3 + r^2 + r + 1 = 0
\]
Factor or find roots:
- Try rational root theorem: possible roots are \( \pm 1 \).
Test \( r = -1 \):
\[
(-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0
\]
Yes! So, \( r = -1 \) is a root.
Factor out \( r + 1 \):
\[
r^3 + r^2 + r + 1 = (r + 1)(r^2 + 0r + 1) = (r + 1)(r^2 + 1)
\]
Solve:
\[
r^2 + 1 = 0 \Rightarrow r = \pm i
\]
Eigenvalues:
\[
r = -1, \quad r = i, \quad r = -i
\]
General Solution of the Annihilated Equation:
\[
Yh = C1 e^{-x} + C2 \cos x + C3 \sin x
\]
---
Step 6: Find the Particular Solution \( Y_p \)
Recall that the original nonhomogeneous term is \( e^{-x} \). The homogeneous solution contains \( e^{-x} \) as well; hence, the particular solution must be modified accordingly to account for duplication (resonance).
Method:
- Since \( e^{-x} \) appears in \( Y_c \), multiply by \( x \) to find a particular solution of the form:
\[
Y_p = A x e^{-x}
\]
- Alternatively, since the annihilator led to solutions involving \( e^{-x} \), the particular solution should be of the form:
\[
Y_p = A x e^{-x}
\]
Determine \( A \):
- Substitute \( Y_p \) into the original differential equation to solve for \( A \).
---
Step 7: Compute Derivatives of \( Y_p \) and Solve for \( A \)
Let:
\[
Y_p = A x e^{-x}
\]
Compute derivatives:
\[
Y_p' = A e^{-x} - A x e^{-x} = A e^{-x} (1 - x)
\]
\[
Y_p'' = -A e^{-x} + A e^{-x} (1 - x) = -A e^{-x} + A e^{-x} - A x e^{-x} = -A x e^{-x}
\]
Now, substitute into the original differential equation:
\[
Y'' - 2Y' + 3Y = e^{-x}
\]
Plugging in:
\[
(-A x e^{-x}) - 2 [A e^{-x} (1 - x)] + 3 [A x e^{-x}] = e^{-x}
\]
Simplify:
\[
-A x e^{-x} - 2A e^{-x} (1 - x) + 3A x e^{-x} = e^{-x}
\]
\[
-A x e^{-x} - 2A e^{-x} + 2A x e^{-x} + 3A x e^{-x} = e^{-x}
\]
Combine like terms:
\[
(-A x + 2A x + 3A x) e^{-x} - 2A e^{-x} = e^{-x}
\]
\[
( ( -A + 2A + 3A ) x - 2A ) e^{-x} = e^{-x}
\]
\[
( (4A) x - 2A ) e^{-x} = e^{-x}
\]
Divide both sides by \( e^{-x} \):
\[
4A x - 2A = 1
\]
Since this must hold for all \( x \), equate coefficients:
- For the \( x \
