Question

using the laplace transform solve y"-3y'-4y=2e^-t when y'(0)=1 and y(0)=1​

Answer 1

The solution of the differential equation is $y(t) = 3e^{(-t)} + 2e^{(4t)}$  as calculated using Laplace transform.

To solve the given differential equation using the Laplace transform, we will first take the Laplace transform of both sides of the equation, apply the initial conditions, and then solve for the Laplace transform of the unknown function. Finally, we will take the inverse Laplace transform to obtain the solution in the time domain.

Given the differential equation:$y" - 3y' - 4y = 2e^{(-t)}$

Step 1: Take the Laplace transform of both sides of the equation.

Taking the Laplace transform of each term:

$L[y"] - 3L[y'] - 4L[y] = 2L[e^{(-t)}]$

Using the properties of the Laplace transform, we have:

$s^2y(s) - sy(0) - y'(0) - 3(sy(s) - y(0)) - 4y(s) = \frac{2}{(s+1)}$

Step 2: Apply the initial conditions.

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

Substituting the initial conditions into the equation, we have:

$s^2y(s) - s - 1 - 3sy(s) + 3 - 4y(s) = \frac{2}{ (s+1)}$

Simplifying further:

$(s^2 - 3s - 4)y(s) - s - 4 = \frac{2} {(s+1)}$

Step 3: Solve for Y(s).

Rearranging the equation:

$(s^2 - 3s - 4)y(s) = \frac{2}{ (s+1)} + s + 4\\(s^2 - 3s - 4)y(s) = \frac{(2 + (s+1)(s+4))} { (s+1)}\\(s^2 - 3s - 4)y(s) = \frac{(s^2 + 5s + 6)} {(s+1)}$

Dividing both sides by ($s^2 - 3s - 4$):

$Y(s) = \frac{(s^2 + 5s + 6)} { [(s+1)(s-4)]}$

Now, we need to express the right side of the equation in terms of partial fractions.

$y(s) = \frac{A} { (s+1)} + \frac{B} {(s-4)}$

Multiplying both sides by [(s+1)(s-4)]:

$s^2 + 5s + 6 = A(s-4) + B(s+1)$

Expanding and comparing coefficients:

$s^2 + 5s + 6 = (A+B)s + (-4A + B)$

Matching the coefficients, we get:

$A + B = 5 ---(1)\\-4A + B = 6 ---(2)$

Solving equations (1) and (2), we find:

$A = 3\\B = 2$

Therefore, the partial fraction decomposition is:

$y(s) = \frac{2} { (s+1)} + \frac{3} {(s-4)}$

Step 4: Take the inverse Laplace transform.

Using the table of Laplace transforms, we can find the inverse Laplace transforms of each term:

$y(t) = 3e^(-t) + 2e^{(4t)}$

Thus, the solution to the given differential equation with the given initial conditions is:

y(t) = 3e^(-t) + 2e^(4t)

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