Consider the initial value problem y" + 36y = cos(6t), y(0) = 4, y' (0) = 2. a. Take the Laplace transform of both sides of the given differential equation to corresponding algebraic equation. Denote the Laplace transform of y(t) by move any terms from one side of the equation to the other (until you get to p s^2Y(s)-4s-2+36Y(s) s/(s^2+36) help (formulas) b. Solve your equation for Y(s). Y(s) = L {y(t)} (((s/(s^2+36))+4s+2)/(s^2+36)) c. Take the inverse Laplace transform of both sides of the previous equation to y(t) = 4cos(6t)+3/2sin(6t)+1/12tsin(6t)
Consider the initial value problem y" + 36y = cos(6t), y(0) = 4, y' (0) = 2. a. Take the Laplace transform of both sides of the given differential equation to corresponding algebraic equation. Denote the Laplace transform of y(t) by move any terms from one side of the equation to the other (until you get to p s^2Y(s)-4s-2+36Y(s) s/(s^2+36) help (formulas) b. Solve your equation for Y(s). Y(s) = L {y(t)} (((s/(s^2+36))+4s+2)/(s^2+36)) c. Take the inverse Laplace transform of both sides of the previous equation to y(t) = 4cos(6t)+3/2sin(6t)+1/12tsin(6t)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Consider the initial value problem**
\[
y'' + 36y = \cos(6t), \quad y(0) = 4, \quad y'(0) = 2.
\]
a. **Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of \( y(t) \) by \( Y(s) \). Do not move any terms from one side of the equation to the other (until you get to part (b) below).**
\[
s^2Y(s) - 4s - 2 + 36Y(s) = \frac{s}{s^2 + 36}
\]
b. **Solve your equation for \( Y(s) \).**
\[
Y(s) = \mathcal{L} \{ y(t) \} = \frac{\left(((s/(s^2+36))+4s+2\right)}{(s^2+36)}
\]
c. **Take the inverse Laplace transform of both sides of the previous equation to solve for \( y(t) \).**
\[
y(t) = 4 \cos(6t) + 3/2 \sin(6t) + 1/12 \sin(6t)
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8204f301-7e54-4f97-9d0f-3395d2e8f77b%2F4ff8fcfd-539f-4d2a-9db2-8fc0c3a991af%2Fivtf7fo_processed.png&w=3840&q=75)
Transcribed Image Text:**Consider the initial value problem**
\[
y'' + 36y = \cos(6t), \quad y(0) = 4, \quad y'(0) = 2.
\]
a. **Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of \( y(t) \) by \( Y(s) \). Do not move any terms from one side of the equation to the other (until you get to part (b) below).**
\[
s^2Y(s) - 4s - 2 + 36Y(s) = \frac{s}{s^2 + 36}
\]
b. **Solve your equation for \( Y(s) \).**
\[
Y(s) = \mathcal{L} \{ y(t) \} = \frac{\left(((s/(s^2+36))+4s+2\right)}{(s^2+36)}
\]
c. **Take the inverse Laplace transform of both sides of the previous equation to solve for \( y(t) \).**
\[
y(t) = 4 \cos(6t) + 3/2 \sin(6t) + 1/12 \sin(6t)
\]
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