## Question:

Question 4(

a)

Compute the Laplace transforms of the following functions.(

i) g(t)=∫0te-2ucos3(t-u)du(

ii) h(t)={3,0≤t<2t2+1,t≥2(

b)

Use Laplace transform to solve the following initial value problem.y”-4y’+3y=8f(t)

with y'(0)=1,y(0)=0

and f(t)={0,0≤t<1e5t,t≥1

## ANSWER:

## Laplace Transforms

**Part (a) – Laplace Transforms**

**(i)**

We can use the following rules of Laplace transform:

**Transform of e^(-at)f(t):**L{e^(-at)f(t)} = F(s+a)**Transform of cos(at):**L{cos(at)} = s/(s^2+a^2)

Therefore,

```
L{g(t)} = L{\int_0^t e^{-2u} cos 3(t-u) du}
```

Using the transformation rule for e^(-at)f(t) with a=2 and F(s) = L{cos(3t)} = 3/(s^2+9), we get:

```
L{g(t)} = e^{-2s} L{cos(3t)} = e^{-2s} * 3/(s^2+9)
```

Applying the transformation rule for cos(at) with a=3:

```
L{g(t)} = e^{-2s} * 3/(s^2+9) = 3e^{-2s} / (s^2+9)
```

**(ii)**

Since h(t) is a piecewise function, we need to find the Laplace transforms of each segment and combine them.

For 0 <= t < 2:

```
L{h(t)} = L{3} = 3/s
```

For t >= 2:

```
L{h(t)} = L{t^2 + 1} = 2/s^3 + 1/s
```

Therefore, the combined Laplace transform of h(t) is:

```
L{h(t)} = {3/s, 0 <= t < 2; 2/s^3 + 1/s, t >= 2}
```

## Initial Value Problem

**Part (b) – Initial Value Problem**

**Taking the Laplace Transform of the equation:**

Apply the Laplace transform to both sides of the given equation:

```
L{y''(t) - 4y'(t) + 3y(t)} = L{8f(t)}
```

Using the properties of the Laplace transform and the initial conditions:

```
s^2 Y(s) - sy(0) - y'(0) - 4sY(s) + 4y(0) + 3Y(s) = 8F(s)
```

Substituting the initial conditions y(0) = 0 and y'(0) = 1:

```
s^2 Y(s) - 1 - 4sY(s) + 3Y(s) = 8F(s)
```

Combining like terms:

```
(s^2 - 4s + 3) Y(s) = 1 + 8F(s)
```

**Solving for Y(s):**

```
Y(s) = (1 + 8F(s)) / (s^2 - 4s + 3)
```

**Finding F(s):**

Since f(t) is piecewise, we need to find the Laplace transforms of each segment and combine them.

For 0 <= t < 1:

```
F(s) = L{0} = 0
```

For t >= 1:

```
F(s) = L{e^5t} = 1 / (s-5)
```

Therefore, the combined Laplace transform of f(t) is:

```
F(s) = {0, 0 <= t < 1; 1/(s-5), t >= 1}
```

**Combining and Solving for Y(s):**

Substituting F(s) and simplifying:

```
Y(s) = (1 + 8(1/(s-5))) / (s^2 - 4s + 3)
```

```
Y(s) = (8/(s-5) + 1) / (s^2 - 4s + 3)
```

**Finding y(t):**

Taking the inverse Laplace transform of Y(s):

```
y(t) = L^{-1}[Y(s)]
```

Using partial fractions and inverse Laplace transforms, we obtain:

```
y(t) = 2e^(3t) - 8e^5t + 1/3, t >= 0
```

Therefore, the solution to the initial value