Answer:
To solve for V(s) in standard form for the given initial value problem, we'll take the Laplace transform of the differential equation. The Laplace transform of a derivative is given by:
L{d^n/dt^n[f(t)]} = s^nF(s) - s^(n-1)f(0) - s^(n-2)f'(0) - ... - f^(n-1)(0)
Where F(s) is the Laplace transform of f(t), and f(0), f'(0), etc., are the initial values of the function and its derivatives.
Given that v(0) = 5 and v'(0) = 10, we have:
v(0) = 5
v'(0) = 10
Now, let's take the Laplace transform of the differential equation:
V(s) = s^2V(s) - s*v(0) - v'(0) + 5sV(s) + 6V(s)
We also have the Laplace transform of the right-hand side:
L{10e^(-t)u(t)} = 10 / (s + 1)
Now, plug in the initial conditions:
V(s) = s^2V(s) - s*5 - 10 + 5sV(s) + 6V(s)
Next, group like terms:
V(s)[s^2 + 5s + 6] = 5s - 10 + 10 / (s + 1)
Now, solve for V(s):
V(s) = [5s - 10 + 10 / (s + 1)] / [s^2 + 5s + 6]
The denominator can be factored:
V(s) = [5s - 10 + 10 / (s + 1)] / [(s + 2)(s + 3)]
Now, we can use partial fraction decomposition to express V(s) in standard form:
V(s) = A/(s + 2) + B/(s + 3) + C/(s + 1)
Multiply both sides by the denominator:
5s - 10 + 10 / (s + 1) = A(s + 3)(s + 1) + B(s + 2)(s + 1) + C(s + 2)(s + 3)
Now, we'll find the values of A, B, and C by equating coefficients of like powers of s on both sides of the equation.
1. Equate the constant terms:
-10 = 3A(1)2 + 2B(1)3 + 2C(3)1
Simplify:
-10 = 3A + 2B + 6C
2. Equate the coefficient of s^1 terms:
5 = 3A(1) + 2B(1) + 2C(1)
Simplify:
5 = 3A + 2B + 2C
3. Equate the coefficient of s^2 terms:
0 = A(1)(3) + B(1)(2) + C(3)(2)
Simplify:
0 = 3A + 2B + 6C
Now, you have a system of three equations with three unknowns:
-10 = 3A + 2B + 6C
5 = 3A + 2B + 2C
0 = 3A + 2B + 6C
You can solve this system of equations to find the values of A, B, and C. Once you have those values, you can express V(s) in standard form.
Explanation: