Question

Use Laplace transforms to solve the following ordinary differential equations:

dtdx​−4x=2e²ᵗ+e⁴ᵗ
Given the initial condition: x(0)=0.

Answer 1

The solution to the given ordinary differential equation using Laplace transforms, with the initial condition x(0) = 0, is:

`\[ x(t) = (1/5)e^(4t) - (1/5)e^(2t) + (2/5)e^(t) \]`

Step-by-step explanation:

To solve the given differential equation
`\[ (dx)/(dt) - 4x = 2e^(2t) + e^(4t) \]`using Laplace transforms, we'll first take the Laplace transform of both sides. Applying the linearity property of Laplace transforms, the transformed equation becomes:

`\[ sX(s) - x(0) - 4X(s) = (2)/(s-2) + (1)/(s-4) \]`

Solving for X(s), we find:

`\[ X(s) = (2)/(s(s-2)(s-4)) + (4)/(s-4) \]`

Now, we can use partial fraction decomposition to simplify X(s) into a sum of simpler fractions. After finding the inverse Laplace transform of each term, we obtain the solution in the time domain:

`\[ x(t) = (1)/(5)e^(4t) - (1)/(5)e^(2t) + (2)/(5)e^(t) \]`

This solution satisfies the initial condition x(0) = 0.

In summary, Laplace transforms provide a powerful tool for solving linear differential equations with constant coefficients. The process involves transforming the differential equation into an algebraic equation in the Laplace domain, solving for the transformed variable, and then finding the inverse Laplace transform to obtain the solution in the time domain.