Question

Use the Laplace transform to solve the given initial-value problem.
y' + 3y = f(t), y(0) = 0
where f(t) = t, 0 ≤ t < 1 0, t ≥ 1

Answer 1

Answer:

The solution to the given Initial - Value - Problem is $y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - [\frac{-1}{9} + \frac{1}{3}t - \frac{2}{9}e^{-3(t-1)}]u(t-1)$

Step-by-step explanation:

y' + 3y = f(t).................(1)

f(t) = t      when 0 ≤ t < 1

f(t) = 0     when t ≥ 1

Step 1: Take the Laplace transform of the LHS of equation (1)

That is L(y' + 3y) = sY(s) + 3Y(s) = Y(s)[s + 3]..............(*)

Step 2: Get an expression for f(t)

For f(t) = t      when 0 ≤ t < 1

f₁(t) = t (1 - u(t - 1)) ( there is a time shift of the unit step)

For f(t) = 0     when t ≥ 1

f₂(t) = 0(u(t-1))

f(t) = f₁(t) + f₂(t)

f(t) = t - t u(t-1)................(2)

Step 3: Taking the Laplace transform of equation (2)

$F(s) = \frac{1}{s^2} - e^{-s} ( \frac{1}{s^2} + \frac{1}{s})$...............(**)

Step 4: Equating * and **

$Y(s) [s + 3]=\frac{1}{s^2} - e^{-s} ( \frac{1}{s^2} + \frac{1}{s}) \\Y(s) = \frac{1}{s^2(s+3)} - e^{-s} ( \frac{1}{s^2(s+3)} + \frac{1}{s(s+3)})$.......................(3)

Since y(t) is the solution we are looking for we need to find the Inverse Laplace Transform of equation (3) by first breaking every  fraction into partial fraction:

$\frac{1}{s^2 (s+3)} = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)}$

$\frac{1}{s (s+3)} = \frac{1}{3s} + \frac{1}{3(s+3)}$

We can rewrite equation (3) by representing the fractions by their partial fractions.

$Y(s) = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} - e^{-s} [\frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} + \frac{1}{3s} + \frac{1}{3(s+3)}]\\Y(s) = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} - e^{-s}[\frac{2}{9s} + \frac{1}{3s^2} - \frac{2}{9(s+3)}]$................(4)

step 5: Take the inverse Laplace transform of equation (4)

$y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - u(t-1)[\frac{2}{9} + \frac{1}{3}(t-1) - \frac{2}{9}e^{-3(t-1)}]$

Simplifying the above equation:

$y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - [\frac{-1}{9} + \frac{1}{3}t - \frac{2}{9}e^{-3(t-1)}]u(t-1)$

Answer 2

The Laplace transform is use to solve the differential equation problem.

The solution for the given initial-value problem is,

$y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)$

Given:

The given initial value problem is $y' + 3y = f(t)$.

Consider the left hand side of the given equation.

$y'+3y$

Take the Laplace transform.

$L(y' + 3y) = sY(s) + 3Y(s) \\L(y' + 3y) = Y(s)[s + 3]$

Consider the right hand side and get the expression for $f(t)$.

$f(t) = t$  when 0 ≤ t < 1

From time shift of the unit step

$f_1(t) = t (1 - u(t - 1))$

For f(t) = 0     when t ≥ 1

Now,

$f_2(t) = 0(u(t-1))f(t) = f_1(t) + f_2(t)f(t) = t - t u(t-1)$

Take the Laplace for above expression.

$F(s)=\dfrac{1}{s^2}-e^{-s}\left(\dfrac{1}{s^2}+\dfrac{1}{s}\right)$

Now, the equate the above two equation.

$Y(s)\left[s+3\right ]=\dfrac{1}{s^2}-e^{-s}\left(\dfrac{1}{s^2}+\dfrac{1}{s}\right)\\Y(s)=\dfrac {1}{(s^2(s+3))}-e^{-s}\left(\dfrac{1}{(s^2(s+3))}+\dfrac{1}{s(s+3)\right)}$

Find the inverse Laplace for the above equation.

$\dfrac{1}{(s^2(s+3))}=\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}\\\dfrac{1}{(s(s+3))}=\dfrac{1}{3s}+\dfrac{1}{3(s+3)}$

Calculate the partial fraction of above equation.

$Y(s)=\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}-e^{-s}\left[\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}+\dfrac{1}{3s}+\dfrac{1}{3(s+3)}\right]\\Y(s)=\dfrac{2}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}-e^{-s}\left[\dfrac{2}{9s}+\dfrac{1}{3s^2}-\dfrac{2}{9(s+3)}\right]$

Take the inverse Laplace of the above equation.

$y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)$

Thus, the solution for the given initial-value problem is,

$y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)$

Learn more about what Laplace transformation is here:

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