find the general solution to linear DE with constant coeffients y'''+y'=2t+3

To find the general solution to the differential equation \( y''' + y' = 2t + 3 \), we will go through the following steps:

### Step 1: Solve the Homogeneous Equation
The corresponding homogeneous equation is:
\[
y''' + y' = 0
\]
This can be simplified by rewriting it as:
\[
D^3 y + D y = 0
\]
where \( D \) represents the differential operator \( \frac{d}{dt} \).

This can be factored as:
\[
D y (D^2 + 1) y = 0
\]
which implies that either \( D y = 0 \) or \( D^2 + 1 = 0 \).

#### Case 1: \( D y = 0 \)
If \( D y = 0 \), then \( y' = 0 \), so \( y = C_1 \), where \( C_1 \) is a constant.

#### Case 2: \( D^2 + 1 = 0 \)
If \( D^2 + 1 = 0 \), then the characteristic equation is:
\[
D^2 = -1
\]
which gives \( D = \pm i \). This results in solutions of the form:
\[
y = C_2 \cos(t) + C_3 \sin(t)
\]
where \( C_2 \) and \( C_3 \) are constants.

Therefore, the general solution to the homogeneous equation is:
\[
y_h = C_1 + C_2 \cos(t) + C_3 \sin(t)
\]

### Step 2: Find a Particular Solution
Now we need to find a particular solution \( y_p \) to the nonhomogeneous equation:
\[
y''' + y' = 2t + 3
\]
Since the right-hand side is a polynomial \( 2t + 3 \), we can try a particular solution of the form:
\[
y_p = At + B
\]
where \( A \) and \( B \) are constants to be determined.

#### Compute \( y_p' \) and \( y_p''' \):
1. \( y_p = At + B \)
2. \( y_p' = A \)
3. \( y_p''' = 0 \)

Substitute into the differential equation:
\[
y''' + y' = 0 + A = 2t + 3
\]
This equation does not balance, so we need to try a higher degree polynomial. We’ll try:
\[
y_p = At^2 + Bt + C
\]

Now, compute the derivatives:
1. \( y_p' = 2At + B \)
2. \( y_p''' = 2A \)

Substitute these into the differential equation:
\[
y''' + y' = 2A + (2At + B) = 2t + 3
\]

Now, match coefficients:
- For the \( t \) term: \( 2A = 2 \Rightarrow A = 1 \)
- For the constant term: \( B + 2A = 3 \Rightarrow B + 2 = 3 \Rightarrow B = 1 \)

So, we have \( A = 1 \) and \( B = 1 \). We can choose any constant \( C \), but it will not affect the particular solution since it corresponds to part of the homogeneous solution.

Thus, the particular solution is:
\[
y_p = t^2 + t
\]

### Step 3: Write the General Solution
The general solution to the differential equation is the sum of the homogeneous solution \( y_h \) and the particular solution \( y_p \):
\[
y = y_h + y_p = C_1 + C_2 \cos(t) + C_3 \sin(t) + t^2 + t
\]

### Final Answer:
\[
y = C_1 + C_2 \cos(t) + C_3 \sin(t) + t^2 + t
\]

Let me know if you would like further details or if you have any questions.

#### Additional Questions to Explore:
1. How would the solution change if the nonhomogeneous term was different, say \( 2t^2 + 3t \)?
2. Can you explain why we tried a second-degree polynomial for the particular solution?
3. What is the significance of the terms in the homogeneous solution?
4. How would the solution method change if we had \( y'' + y = 2t + 3 \) instead?
5. Can we apply variation of parameters here, and would it yield the same result?

#### Tip:
When guessing a particular solution, always try a form similar to the nonhomogeneous term, adjusting the degree if necessary.

Find the general solution to the linear differential equation with constant coefficients y''' + y' = 2t + 3

Learn how to solve the linear differential equation y''' + y' = 2t + 3 with constant coefficients. This solution explains step-by-step methods, including solving the homogeneous equation, finding a particular solution, and combining these for the general solution. Ideal for high school and college students studying 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