 Solve the O.D.E.
y ′′ − 4y ′ + 4y = 5 cos(t)


The given second-order linear non-homogeneous ordinary differential equation is:

\[
y'' - 4y' + 4y = 5\cos(t)
\]

To solve this, we will proceed in two steps:
1. Solve the corresponding **homogeneous equation**: \( y'' - 4y' + 4y = 0 \).
2. Find a **particular solution** for the non-homogeneous equation.

### Step 1: Solve the Homogeneous Equation

The homogeneous equation is:
\[
y'' - 4y' + 4y = 0
\]
This is a linear differential equation with constant coefficients. To solve it, we assume a solution of the form \( y_h(t) = e^{rt} \), where \( r \) is a constant to be determined.

Substitute \( y_h(t) = e^{rt} \) into the homogeneous equation:
\[
r^2 e^{rt} - 4r e^{rt} + 4 e^{rt} = 0
\]
Factor out \( e^{rt} \) (which is never zero):
\[
e^{rt}(r^2 - 4r + 4) = 0
\]
So, we have the characteristic equation:
\[
r^2 - 4r + 4 = 0
\]
This simplifies to:
\[
(r - 2)^2 = 0
\]
Thus, \( r = 2 \) is a **repeated root**. Therefore, the general solution to the homogeneous equation is:
\[
y_h(t) = (C_1 + C_2 t) e^{2t}
\]
where \( C_1 \) and \( C_2 \) are arbitrary constants.

### Step 2: Find a Particular Solution

Now, we need to find a particular solution to the non-homogeneous equation:
\[
y'' - 4y' + 4y = 5\cos(t)
\]

Since the right-hand side is \( 5\cos(t) \), we guess a particular solution of the form:
\[
y_p(t) = A\cos(t) + B\sin(t)
\]
where \( A \) and \( B \) are constants to be determined.

Now, compute the first and second derivatives of \( y_p(t) \):
\[
y_p'(t) = -A\sin(t) + B\cos(t)
\]
\[
y_p''(t) = -A\cos(t) - B\sin(t)
\]

Substitute \( y_p(t) \), \( y_p'(t) \), and \( y_p''(t) \) into the original differential equation:
\[
(-A\cos(t) - B\sin(t)) - 4(-A\sin(t) + B\cos(t)) + 4(A\cos(t) + B\sin(t)) = 5\cos(t)
\]

Simplifying the left-hand side:
\[
-A\cos(t) - B\sin(t) + 4A\sin(t) - 4B\cos(t) + 4A\cos(t) + 4B\sin(t)
\]
Group the terms involving \( \cos(t) \) and \( \sin(t) \):
\[
(A\cos(t)) + (3B\sin(t)) = 5\cos(t)
\]

Now, compare the coefficients of \( \cos(t) \) and \( \sin(t) \) on both sides of the equation. For the coefficients of \( \cos(t) \):
\[
A = 5
\]
And for the coefficients of \( \sin(t) \):
\[
3B = 0 \quad \Rightarrow \quad B = 0
\]

Thus, the particular solution is:
\[
y_p(t) = 5\cos(t)
\]

### Step 3: General Solution

The general solution to the non-homogeneous equation is the sum of the homogeneous and particular solutions:
\[
y(t) = y_h(t) + y_p(t) = (C_1 + C_2 t)e^{2t} + 5\cos(t)
\]

### Final Answer:
\[
y(t) = (C_1 + C_2 t)e^{2t} + 5\cos(t)
\]

---

Would you like further clarification on any step, or do you have other questions? 

Here are some related questions:
1. How do we solve a non-homogeneous ODE with a sine or cosine term?
2. What if the right-hand side of the ODE was a sine function instead of a cosine?
3. How do you find the general solution to an ODE with repeated roots in the characteristic equation?
4. What does the constant \( C_1 \) and \( C_2 \) represent in the solution?
5. How would the solution change if the non-homogeneous term was an exponential function?

**Tip**: Always check the form of the non-homogeneous term when choosing an ansatz for a particular solution!

Solve the O.D.E. y ′′ − 4y ′ + 4y = 5 cos(t)

Learn how to solve the second-order linear non-homogeneous differential equation y ′′ − 4y ′ + 4y = 5 cos(t) with detailed step-by-step solutions. This problem explores methods like solving homogeneous equations and finding particular solutions using the method of undetermined coefficients. Suitable for advanced high school and university students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation