https://file.aiquickdraw.com/mathbot/file/1733883065506r53y5wuu.jpg

Let's solve question (d) from the image using Laplace transforms.

The equation is:

\[
F'(t) + 4F(t) = \sin 2t, \quad F(0) = 1, \quad F'(0) = -2
\]

### Step 1: Take the Laplace transform
The Laplace transform of the given equation is:

\[
L\{F'(t)\} + 4L\{F(t)\} = L\{\sin 2t\}
\]

Using Laplace transform properties:
- \( L\{F'(t)\} = sF(s) - F(0) \)
- \( L\{F(t)\} = F(s) \)
- \( L\{\sin 2t\} = \frac{2}{s^2 + 4} \)

Substitute these into the equation:

\[
[sF(s) - F(0)] + 4F(s) = \frac{2}{s^2 + 4}
\]

Given \( F(0) = 1 \), this becomes:

\[
sF(s) - 1 + 4F(s) = \frac{2}{s^2 + 4}
\]

### Step 2: Combine terms
Factor out \( F(s) \) on the left-hand side:

\[
F(s)(s + 4) = \frac{2}{s^2 + 4} + 1
\]

Simplify the right-hand side by writing \( 1 \) with a common denominator:

\[
\frac{2}{s^2 + 4} + 1 = \frac{2 + (s^2 + 4)}{s^2 + 4} = \frac{s^2 + 6}{s^2 + 4}
\]

Thus, the equation becomes:

\[
F(s)(s + 4) = \frac{s^2 + 6}{s^2 + 4}
\]

Solve for \( F(s) \):

\[
F(s) = \frac{s^2 + 6}{(s + 4)(s^2 + 4)}
\]

### Step 3: Perform partial fraction decomposition
Decompose \( \frac{s^2 + 6}{(s + 4)(s^2 + 4)} \):

\[
\frac{s^2 + 6}{(s + 4)(s^2 + 4)} = \frac{A}{s + 4} + \frac{Bs + C}{s^2 + 4}
\]

Multiply through by \( (s + 4)(s^2 + 4) \) to get:

\[
s^2 + 6 = A(s^2 + 4) + (Bs + C)(s + 4)
\]

Expand both terms:
- \( A(s^2 + 4) = As^2 + 4A \)
- \( (Bs + C)(s + 4) = Bs^2 + 4Bs + Cs + 4C = Bs^2 + (4B + C)s + 4C \)

Combine:

\[
s^2 + 6 = (A + B)s^2 + (4B + C)s + (4A + 4C)
\]

Equating coefficients:
1. For \( s^2 \): \( A + B = 1 \)
2. For \( s \): \( 4B + C = 0 \)
3. For the constant: \( 4A + 4C = 6 \)

### Step 4: Solve for \( A \), \( B \), and \( C \)
From \( A + B = 1 \), \( B = 1 - A \).

Substitute \( B = 1 - A \) into \( 4B + C = 0 \):

\[
4(1 - A) + C = 0 \implies 4 - 4A + C = 0 \implies C = 4A - 4
\]

Substitute \( A \) and \( C \) into \( 4A + 4C = 6 \):

\[
4A + 4(4A - 4) = 6 \implies 4A + 16A - 16 = 6 \implies 20A = 22 \implies A = \frac{11}{10}
\]

Solve for \( B \) and \( C \):
- \( B = 1 - A = 1 - \frac{11}{10} = -\frac{1}{10} \)
- \( C = 4A - 4 = 4\left(\frac{11}{10}\right) - 4 = \frac{44}{10} - \frac{40}{10} = \frac{4}{10} = \frac{2}{5} \)

Thus:

\[
\frac{s^2 + 6}{(s + 4)(s^2 + 4)} = \frac{\frac{11}{10}}{s + 4} + \frac{-\frac{1}{10}s + \frac{2}{5}}{s^2 + 4}
\]

### Step 5: Take the inverse Laplace transform
Write \( F(s) \) as:

\[
F(s) = \frac{\frac{11}{10}}{s + 4} + \frac{-\frac{1}{10}s}{s^2 + 4} + \frac{\frac{2}{5}}{s^2 + 4}
\]

The inverse Laplace transforms are:
1. \( L^{-1}\left\{\frac{1}{s + 4}\right\} = e^{-4t} \)
2. \( L^{-1}\left\{\frac{s}{s^2 + 4}\right\} = \cos(2t) \)
3. \( L^{-1}\left\{\frac{1}{s^2 + 4}\right\} = \frac{\sin(2t)}{2} \)

Using linearity:

\[
F(t) = \frac{11}{10}e^{-4t} - \frac{1}{10}\cos(2t) + \frac{2}{5} \cdot \frac{\sin(2t)}{2}
\]

Simplify:

\[
F(t) = \frac{11}{10}e^{-4t} - \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t)
\]

### Final Answer:

\[
F(t) = \frac{11}{10}e^{-4t} - \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t)
\]

Would you like me to explain any step in more detail?  

### Follow-up Questions:
1. What is the purpose of Laplace transforms in solving differential equations?
2. Why is partial fraction decomposition needed in this problem?
3. Can you explain how the coefficients \( A, B, C \) were determined?
4. What are the key properties of the Laplace transform used in this solution?
5. How would the solution change if initial conditions were different?

### Tip:
When solving Laplace transform problems, always simplify expressions as much as possible before applying partial fraction decomposition.

Learn how to solve a first-order differential equation with Laplace transforms, covering step-by-step solutions including Laplace properties, partial fractions, and inverse transforms. Suitable for undergraduate mathematics or engineering students.

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