Question

Using Laplace translotms techinique, solve given y′′−2y′+2y=0 Solve the IVP : x′′+4x=8δ2π​(t),x(0)=3,x′(0)=0 using LT. Find the general solution to x₁′​=6x₁+x₂+6t′,x₂′​=4x₁​+3x₂​−

Answer 1

By solving these equations for
`\(x_1\) and \(x_2\)` in terms of
`\(t\),` the expressions
`\(x_1(t)\) and \(x_2(t)\)`are derived as functions of
`\(t\)`.
`\(x_1(t) = 4e^(-2t) + 2e^(2t) + 4t - 1\), \(x_2(t) = 2e^(-2t) - 2e^(2t) + 2t^2 - 1\)`

Step-by-step explanation:

The problem requires solving a system of differential equations using Laplace transforms. Firstly, the Laplace transform of the given system of equations is obtained. Then, after solving the resulting algebraic equations, the inverse Laplace transform is applied to find the solutions for
`\(x_1(t)\) and \(x_2(t)`\).

The Laplace transform technique involves converting the differential equations into algebraic equations in the Laplace domain. For the given system of equations
`\(x_1' = 6x_1 + x_2 + 6t'\) and \(x_2' = 4x_1 + 3x_2\),`taking the Laplace transforms and rearranging terms lead to a system of algebraic equations.

By solving these equations for
`\(x_1\) and \(x_2\)` in terms of
`\(t\),` the expressions
`\(x_1(t)\) and \(x_2(t)\)`are derived as functions of
`\(t\)`. The solutions are \
`(x_1(t) = 4e^(-2t) + 2e^(2t) + 4t - 1\) and \(x_2(t) = 2e^(-2t) - 2e^(2t) + 2t^2 - 1\).`

This demonstrates how the Laplace transform method can be applied to solve systems of differential equations, providing explicit solutions in terms of the given functions and variables.