Question

What is the Laplace transform \( X(s) \) of the solution of the differential equation \[ 2 \frac{d^{2}}{d t^{2}} x-2 \frac{d}{d t} x+10 x=\cos (3 t)+\mathrm{e}^{t} \] with initial conditions \( x(0)=0\ and x ′

(0)=1 ? Give your answer in the form X=p(s), where p(s) is a rational function of s, that is, your answer need to start with " X= ".

Answer 1

To find the Laplace transform of the solution \(X(s)\) for the given differential equation, we will apply the initial conditions and the properties of the Laplace transform.

1. First, let's write down the Laplace transform of the left-hand side of the differential equation:

Remember that the Laplace transform of the \(n\)th derivative of \(x\) is \(s^n\cdot X(s) - s^{n-1}x(0) - s^{n-2}x'(0) - \ldots - x^{(n-1)}(0)\). Applying this to the left side of our equation with \(n=2\) and using the given initial conditions we get:

\(L[2x'' - 2x' + 10x] = L[2x''] - L[2x'] + L[10x] = 2s^2X - 2s*0 - 2*1 - 2sX + 2*0 + 10X = 2s^2X -2 + 10X - 2sX = 2(s^2-s)X - 2 + 10X.\)

2. Second, let's calculate the Laplace transform of the right-hand side of the differential equation:

The Laplace transform of \(\cos(at)\) is \(s/(s^2 + a^2)\) and the Laplace transform of \(e^{at}\) is \(1/(s-a)\). Using these results, we have:

\(L[\cos(3t) + e^t] = L[\cos(3t)] + L[e^t] = s/(s^2 + 9) + 1/(s-1).\)

3. Equating the Laplace transforms of both sides of the equation:

\(2(s^2-s)X -2 + 10X = s/(s^2 + 9) + 1/(s-1).\)

4. Solving this equation for \(X\), we can simplify it as:

\(X = (s/(2s^2 + 18) + 1/(2s-2) + 1)/((s^2-s) + 5).\)

So, this is the Laplace transform of the solution for the differential equation given the provided initial conditions.