To solve the initial value problem using the Laplace transform, we start by applying the Laplace transform to both sides of the equation.
The given first-order linear differential equation is:
y’ = 5y + e^(4t)
Taking the Laplace transform of the left-hand side, we have:
L{y’} = sY(s) – y(0)
where Y(s) is the Laplace transform of y(t) and y(0) = 2 as given in the problem.
Substituting this into the equation gives:
sY(s) – 2 = 5Y(s) + L{e^(4t)}
Next, we find the Laplace transform of e^(4t), which is:
L{e^(4t)} = rac{1}{s – 4}
Now substituting back, we have:
sY(s) – 2 = 5Y(s) + rac{1}{s – 4}
Rearranging the terms gives:
sY(s) – 5Y(s) = 2 + rac{1}{s – 4}
This simplifies to:
(s – 5)Y(s) = 2 + rac{1}{s – 4}
Now, solving for Y(s), we have:
Y(s) = rac{2 + rac{1}{s – 4}}{s – 5}
This can be rewritten as:
Y(s) = rac{2(s – 5) + 1}{(s – 5)(s – 4)}
Y(s) = rac{2s – 10 + 1}{(s – 5)(s – 4)} = rac{2s – 9}{(s – 5)(s – 4)}
Next, we will perform partial fraction decomposition:
rac{2s – 9}{(s – 5)(s – 4)} = rac{A}{s – 5} + rac{B}{s – 4}
Multiplying through by the denominator gives:
2s – 9 = A(s – 4) + B(s – 5)
To find constants A and B, we can choose convenient values for s:
Let s = 5:
2(5) – 9 = A(5 – 4) + B(0) → 10 – 9 = A(1) → A = 1
Let s = 4:
2(4) – 9 = A(0) + B(4 – 5) → 8 – 9 = B(-1) → B = 1
So we have:
Y(s) = rac{1}{s – 5} + rac{1}{s – 4}
The inverse Laplace transform gives:
y(t) = L^{-1}igg{ rac{1}{s – 5}igg} + L^{-1}igg{ rac{1}{s – 4}igg}
This means:
y(t) = e^{5t} + e^{4t} + C
Where C is a constant determined by initial conditions. We already know y(0) = 2. Thus:
y(0) = e^{5(0)} + e^{4(0)} + C = 2 → 1 + 1 + C = 2
This simplifies to C = 0. Therefore, the final solution is:
y(t) = e^{5t} + e^{4t}
In conclusion, we used the Laplace transform to solve the differential equation and found the solution to be y(t) = e^{5t} + e^{4t}.