Question

How do I solve this problem using Laplace transform? y'-3y=e^2x y(0)=2

Answer 1

The solution to the given differential equation using Laplace transform is
`\( y(t) = e^(3t) + (e^(2t))/(5) \)`.

Step-by-step explanation:

In order to solve the differential equation
`\(y' - 3y = e^(2t)\)` with the initial condition
`\(y(0) = 2\)` using Laplace transform, we first take the Laplace transform of both sides of the equation. The Laplace transform of the derivative
`\(y'\) is \(sY(s) - y(0)\)`, where
`\(Y(s)\)` is the Laplace transform of
`\(y(t)\)` and
`\(s\)` is the Laplace variable. Applying this, the Laplace transform of the given equation becomes
`\(sY(s) - 2 - 3Y(s) = (1)/(s - 2)\)`.

Next, we solve for
`\(Y(s)\)` and then take the inverse Laplace transform to find
`\(y(t)\)`. After rearranging terms and solving, we get
`\(Y(s) = (s - 2)/((s - 3)(s - 2))\)`. Partial fraction decomposition is then applied to express
`\(Y(s)\)` in a form that allows for an easy inverse Laplace transform. The result is
`\(Y(s) = (1)/(s - 3) + (1)/(5(s - 2))\)`.

Finally, taking the inverse Laplace transform of
`\(Y(s)\)` gives the solution
`\(y(t) = e^(3t) + (e^(2t))/(5)\)`. The initial condition
`\(y(0) = 2\)` is satisfied, and this solution represents the solution to the given differential equation.