Solve the initial value problem using the method of Laplace transforms: $y^{\mathrm{\text{'}\text{'}}}+4y=0$, $y\left(0\right)=2$, $y^{\mathrm{\text{'}}}\left(0\right)=0$. What is $y\left(t\right)$?

• A. $y\left(t\right)=2e^{2t}$
• B. $y\left(t\right)=2\sin \left(2t\right)$
• C. $y\left(t\right)=2e^{-2t}$
• D. $y\left(t\right)=2\cos \left(2t\right)$

Answer: (D)

Verified step by step guidance

1. Step 1: Apply the Laplace transform to the given differential equation y'' + 4y = 0. Recall that the Laplace transform of y'' is \( \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) \), and the Laplace transform of y is \( \mathcal{L}\{y\} = Y(s) \). Substitute the initial conditions y(0) = 2 and y'(0) = 0 into the transformed equation.
2. Step 2: After applying the Laplace transform, the equation becomes \( s^2Y(s) - 2 + 4Y(s) = 0 \). Rearrange this equation to solve for \( Y(s) \), the Laplace transform of the solution y(t).
3. Step 3: Simplify the equation to isolate \( Y(s) \). Combine terms to get \( Y(s)(s^2 + 4) = 2 \), and then divide through by \( s^2 + 4 \) to find \( Y(s) = \frac{2}{s^2 + 4} \).
4. Step 4: Recognize that \( \frac{2}{s^2 + 4} \) is the Laplace transform of \( 2 \cos(2t) \). Use the inverse Laplace transform to convert \( Y(s) \) back into the time domain, yielding \( y(t) = 2 \cos(2t) \).
5. Step 5: Verify the solution by substituting \( y(t) = 2 \cos(2t) \) back into the original differential equation y'' + 4y = 0 and checking that it satisfies the initial conditions y(0) = 2 and y'(0) = 0.