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.