Find a particular solution for: y’’+0y’-1y=2t^2+1t-5e^-4t

To find a particular solution to the differential equation:

\[
y'' - y = 2t^2 + t - 5e^{-4t},
\]

we'll proceed by using the method of undetermined coefficients. We will find a particular solution \( y_p \) that satisfies the non-homogeneous equation.

### Step 1: Identify the form of the particular solution

Since the right-hand side consists of a polynomial term \( 2t^2 + t \) and an exponential term \( -5e^{-4t} \), we can assume the particular solution as:

\[
y_p = y_p^{(1)} + y_p^{(2)},
\]

where:
- \( y_p^{(1)} \) is the particular solution for the polynomial part \( 2t^2 + t \),
- \( y_p^{(2)} \) is the particular solution for the exponential part \( -5e^{-4t} \).

### Step 2: Guess for the polynomial part \( 2t^2 + t \)

Since the non-homogeneous term \( 2t^2 + t \) is a quadratic polynomial, we assume a particular solution of the form:

\[
y_p^{(1)} = At^2 + Bt + C.
\]

### Step 3: Differentiate \( y_p^{(1)} \) and substitute

1. Compute the derivatives:
   \[
   y_p^{(1)'} = 2At + B,
   \]
   \[
   y_p^{(1)''} = 2A.
   \]

2. Substitute into the differential equation:
   \[
   y_p^{(1)''} - y_p^{(1)} = 2A - (At^2 + Bt + C).
   \]

   Equating this to \( 2t^2 + t \), we get:
   \[
   2A - At^2 - Bt - C = 2t^2 + t.
   \]

3. Equating coefficients, we have:
   - From \( t^2 \): \( -A = 2 \) ⟹ \( A = -2 \),
   - From \( t \): \( -B = 1 \) ⟹ \( B = -1 \),
   - From the constant term: \( 2A - C = 0 \) ⟹ \( 2(-2) - C = 0 \) ⟹ \( C = -4 \).

Thus, the particular solution for the polynomial part is:
\[
y_p^{(1)} = -2t^2 - t - 4.
\]

### Step 4: Guess for the exponential part \( -5e^{-4t} \)

Since the non-homogeneous term includes \( -5e^{-4t} \), we assume a particular solution of the form:

\[
y_p^{(2)} = De^{-4t}.
\]

### Step 5: Differentiate \( y_p^{(2)} \) and substitute

1. Compute the derivatives:
   \[
   y_p^{(2)'} = -4De^{-4t},
   \]
   \[
   y_p^{(2)''} = 16De^{-4t}.
   \]

2. Substitute into the differential equation:
   \[
   y_p^{(2)''} - y_p^{(2)} = 16De^{-4t} - De^{-4t} = 15De^{-4t}.
   \]

   Equating this to \( -5e^{-4t} \), we get:
   \[
   15D = -5 \quad \Rightarrow \quad D = -\frac{1}{3}.
   \]

Thus, the particular solution for the exponential part is:
\[
y_p^{(2)} = -\frac{1}{3}e^{-4t}.
\]

### Step 6: Combine the results

The particular solution is:
\[
y_p = y_p^{(1)} + y_p^{(2)} = -2t^2 - t - 4 - \frac{1}{3}e^{-4t}.
\]

### Final Answer:

\[
\boxed{y_p = -2t^2 - t - 4 - \frac{1}{3}e^{-4t}.}
\]

Would you like more details on any of these steps or have any questions?

### Relative Questions

1. How would you find the general solution of this differential equation?
2. What is the method of undetermined coefficients and when is it used?
3. Can we use another method like variation of parameters to solve this problem?
4. How do you handle different types of non-homogeneous terms when finding particular solutions?
5. What changes if one of the terms on the right-hand side is a repeated root in the characteristic equation?

### Tip:
Always check if your assumed particular solution overlaps with the homogeneous solution. If it does, multiply the assumed solution by \( t \) until there is no overlap.

Find a particular solution for: y’’ + 0y’ - y = 2t^2 + 1t - 5e^-4t

Explore how to find a particular solution for the non-homogeneous differential equation y'' - y = 2t^2 + t - 5e^-4t using the method of undetermined coefficients. The detailed solution involves solving polynomial and exponential terms step-by-step, suitable for students in Grades 11-12 or early college.

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