Question

Let \(y(t)\) be the solution of the differential equation \(y''+4y=0\), \(y(0)=1\), \(y'(0)=-6\), then the Laplace transformation \(Y(s)\) of the solution is equal to

• A. \( \frac{s}{s^2+4} + \frac{2}{s^2+4} \)
• B. \( \frac{s-6}{s^2-4} \)
• C. \( \frac{s+6}{s^2+4} \)
• D. \( \frac{s-6}{s^2+4} \)

Hint:

The key formula for solving initial value problems with Laplace transforms is: \( \mathcal{L}\{y^{(n)}\} = s^n Y(s) - s^{n-1}y(0) - s^{n-2}y'(0) - \dots - y^{(n-1)}(0) \).

Answer 1

The Correct Option is D

Step 1: Take the Laplace transform of the entire differential equation. \[ \mathcal{L}\{y''\} + \mathcal{L}\{4y\} = \mathcal{L}\{0\} \] \[ \mathcal{L}\{y''\} + 4\mathcal{L}\{y\} = 0 \]

Step 2: Apply the formula for the Laplace transform of derivatives. Let \(Y(s) = \mathcal{L}\{y(t)\}\). The formula is \( \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) \). Substitute this into the transformed equation: \[ (s^2Y(s) - sy(0) - y'(0)) + 4Y(s) = 0 \]

Step 3: Substitute the given initial conditions \(y(0)=1\) and \(y'(0)=-6\). \[ s^2Y(s) - s(1) - (-6) + 4Y(s) = 0 \] \[ s^2Y(s) - s + 6 + 4Y(s) = 0 \]

Step 4: Solve for \(Y(s)\). \[ (s^2+4)Y(s) = s-6 \] \[ Y(s) = \frac{s-6}{s^2+4} \] This can also be written as \( \frac{s}{s^2+4} - \frac{6}{s^2+4} \).