find the general solution of the non-homogenous ode y''(t) -y(t) = cosh(t)

To solve the non-homogeneous ordinary differential equation (ODE):

\[
y''(t) - y(t) = \cosh(t)
\]

we follow these steps:

### Step 1: Solve the corresponding homogeneous equation
The homogeneous equation is:

\[
y''(t) - y(t) = 0
\]

The characteristic equation associated with this ODE is:

\[
r^2 - 1 = 0
\]

This factors as:

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

Thus, the roots are \( r = 1 \) and \( r = -1 \).

The general solution to the homogeneous equation is:

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

where \( c_1 \) and \( c_2 \) are arbitrary constants.

### Step 2: Find a particular solution to the non-homogeneous equation
The non-homogeneous term is \( \cosh(t) \). Recall that:

\[
\cosh(t) = \frac{e^t + e^{-t}}{2}
\]

We look for a particular solution, \( y_p(t) \), in the form of:

\[
y_p(t) = A e^t + B e^{-t}
\]

Taking the derivatives:

\[
y_p'(t) = A e^t - B e^{-t}
\]
\[
y_p''(t) = A e^t + B e^{-t}
\]

Substituting \( y_p(t) \), \( y_p'(t) \), and \( y_p''(t) \) into the original equation:

\[
(A e^t + B e^{-t}) - (A e^t + B e^{-t}) = \frac{e^t + e^{-t}}{2}
\]

Simplifying, this becomes:

\[
0 = \frac{e^t + e^{-t}}{2}
\]

Since this is not possible (the left-hand side is always 0 while the right-hand side is non-zero), this implies that our particular solution assumption must be revised.

Because the non-homogeneous term \( \cosh(t) \) is already a linear combination of the solutions to the homogeneous equation, we multiply our guess by \( t \) to get:

\[
y_p(t) = At e^t + Bt e^{-t}
\]

Repeating the process by differentiating and substituting into the original ODE, we find:

\[
y_p''(t) - y_p(t) = (At e^t + Bt e^{-t} + Ae^t - Be^{-t}) - (At e^t + Bt e^{-t}) = (A e^t - B e^{-t}) 
\]

Equating the coefficients to those of \( \cosh(t) \):

\[
A e^t - B e^{-t} = \frac{e^t + e^{-t}}{2}
\]

This yields:

\[
A = \frac{1}{2}, \quad B = -\frac{1}{2}
\]

Thus, the particular solution is:

\[
y_p(t) = \frac{1}{2} t e^t - \frac{1}{2} t e^{-t}
\]

### Step 3: Write the general solution
The general solution of the ODE is the sum of the homogeneous and particular solutions:

\[
y(t) = y_h(t) + y_p(t) = c_1 e^t + c_2 e^{-t} + \frac{1}{2} t e^t - \frac{1}{2} t e^{-t}
\]

### Final Answer:
\[
\boxed{y(t) = \left(c_1 + \frac{1}{2} t \right) e^t + \left(c_2 - \frac{1}{2} t \right) e^{-t}}
\]

Would you like further details on any of the steps?

### Related Questions:
1. How do you solve a homogeneous second-order differential equation?
2. What are the methods for solving non-homogeneous ODEs?
3. How does the method of undetermined coefficients work?
4. Can you explain the Wronskian and its significance in differential equations?
5. What are the differences between homogeneous and non-homogeneous equations?
6. How can we apply variation of parameters to solve ODEs?
7. What are the common forms of particular solutions for specific right-hand sides in ODEs?
8. Can you explain the significance of characteristic equations in solving ODEs?

**Tip:** Always verify the solution by substituting it back into the original differential equation. This ensures that the solution satisfies the equation.

Explore the detailed solution to the non-homogeneous ordinary differential equation (ODE) y''(t) - y(t) = cosh(t). Learn how to solve using the characteristic equation and the method of undetermined coefficients. Suitable for advanced undergraduate students and enthusiasts of differential equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation