Question

1. Find Laplace transform of ?(?) = 2 + 5?

2. Find Laplace transform of ?(?) = 2?^-t + 3??^-4t

3. Find time function corresponding to this Laplace transform: ?(?) = (2s²+s+1)/(s³-1)

4. Solve this ODE using Laplace transform : ?̈(?)+2?̇(?)+4?(?)=0; ?(0)=1, ?̇(0)=2

5. Solve this ODE using the Laplace transform : ?̈(?)−2?̇(?)+3?(?)=0; ?(0)=2, ?̇(0)=1

Answer 1

Answer 2

To find the Laplace transforms of the given functions, we'll use the standard formulas and properties of Laplace transforms.

1. Laplace transform of f(t) = 2 + 5t: The Laplace transform of a constant is simply the constant multiplied by the Laplace transform of 1, which is 1/s. The Laplace transform of t is 1/s^2. Therefore, applying linearity, the Laplace transform of 2 + 5t is: L{2 + 5t} = 2 * L{1} + 5 * L{t} = 2 * 1/s + 5 * 1/s^2 = (2/s) + (5/s^2)

2. Laplace transform of f(t) = 2e^(-t) + 3t^2e^(-4t): We'll use the linearity and the exponential shift property of Laplace transforms. The Laplace transform of e^(-at) is 1/(s + a). Applying linearity, the Laplace transform of 2e^(-t) is 2 * L{e^(-t)} = 2/(s + 1). For 3t^2e^(-4t), we differentiate twice with respect to t to get the transform. The Laplace transform of t^2 is 2!/s^3, and the Laplace transform of e^(-4t) is 1/(s + 4). Applying linearity, the Laplace transform of 3t^2e^(-4t) is 3 * (2!/s^3) * (1/(s + 4)) = 6/(s^3(s + 4)). Adding the two terms together, we have: L{2e^(-t) + 3t^2e^(-4t)} = 2/(s + 1) + 6/(s^3(s + 4))

3. Finding the time function corresponding to ?(s) = (2s^2 + s + 1)/(s^3 - 1): To find the inverse Laplace transform of this function, we need to factor the denominator of the rational function and use partial fractions. The denominator factors as (s - 1)(s^2 + s + 1). The roots of the quadratic factor can be found using the quadratic formula as (-1 ± sqrt(3)i)/2. We'll write the partial fractions as A/(s - 1) + (Bs + C)/(s^2 + s + 1), where A, B, and C are constants to be determined.

Using the method of partial fractions, we can solve for A, B, and C by equating numerators: 2s^2 + s + 1 = A(s^2 + s + 1) + (Bs + C)(s - 1)

Expanding the right side and equating coefficients, we get the following system of equations: 2 = A + B 1 = A - B + C 1 = A - C

Solving this system of equations, we find A = 1, B = 1, and C = 0.

Therefore, the time function corresponding to ?(s) = (2s^2 + s + 1)/(s^3 - 1) is: f(t) = L^(-1){?(s)} = L^(-1){1/(s - 1)} + L^(-1){(s + 1)/(s^2 + s + 1)} = e^t + L^(-1){(s + 1