Math Problem Statement

solve

Use the Laplace transform to find solution y of the differential equation: y'' + 9y = 0, y(0) = y0, y'(0) = y1

Solution

To solve this differential equation using the Laplace transform, let us break it down step-by-step:

Problem Statement

We are solving y+9y=0y'' + 9y = 0, with initial conditions:

  • y(0)=y0y(0) = y_0,
  • y(0)=y1y'(0) = y_1.

Step 1: Apply the Laplace transform

Take the Laplace transform of both sides of the equation. Recall the Laplace transform rules:

  1. L(y(t))=sY(s)y(0)\mathcal{L}(y'(t)) = sY(s) - y(0),
  2. L(y(t))=s2Y(s)sy(0)y(0)\mathcal{L}(y''(t)) = s^2Y(s) - sy(0) - y'(0).

Applying this, we get: L(y)+9L(y)=0.\mathcal{L}(y'') + 9\mathcal{L}(y) = 0.

Substituting the Laplace transforms: s2Y(s)sy(0)y(0)+9Y(s)=0.s^2Y(s) - sy(0) - y'(0) + 9Y(s) = 0.

Simplify using the initial conditions y(0)=y0y(0) = y_0 and y(0)=y1y'(0) = y_1: s2Y(s)sy0y1+9Y(s)=0.s^2Y(s) - sy_0 - y_1 + 9Y(s) = 0.

Step 2: Solve for Y(s)Y(s)

Group terms involving Y(s)Y(s): Y(s)(s2+9)=sy0+y1.Y(s)(s^2 + 9) = sy_0 + y_1.

Solve for Y(s)Y(s): Y(s)=sy0+y1s2+9.Y(s) = \frac{sy_0 + y_1}{s^2 + 9}.

Step 3: Take the inverse Laplace transform

To find y(t)y(t), apply the inverse Laplace transform to Y(s)Y(s). First, rewrite Y(s)Y(s) as: Y(s)=sy0s2+9+y1s2+9.Y(s) = \frac{sy_0}{s^2 + 9} + \frac{y_1}{s^2 + 9}.

Using the standard Laplace transform pairs:

  1. L1(ss2+ω2)=cos(ωt)\mathcal{L}^{-1}\left(\frac{s}{s^2 + \omega^2}\right) = \cos(\omega t),
  2. L1(1s2+ω2)=sin(ωt)ω\mathcal{L}^{-1}\left(\frac{1}{s^2 + \omega^2}\right) = \frac{\sin(\omega t)}{\omega},

we compute: L1(sy0s2+9)=y0cos(3t),\mathcal{L}^{-1}\left(\frac{sy_0}{s^2 + 9}\right) = y_0 \cos(3t), L1(y1s2+9)=y13sin(3t).\mathcal{L}^{-1}\left(\frac{y_1}{s^2 + 9}\right) = \frac{y_1}{3} \sin(3t).

Thus, the solution is: y(t)=y0cos(3t)+y13sin(3t).y(t) = y_0 \cos(3t) + \frac{y_1}{3} \sin(3t).

Step 4: Verify the answer

From the solution, we see that the correct answer is D: y(t)=y0cos(3t)+y13sin(3t).y(t) = y_0 \cos(3t) + \frac{y_1}{3} \sin(3t).


Would you like additional clarification or detailed explanations of any step?

Related Questions:

  1. How do you handle initial conditions in Laplace transforms for higher-order equations?
  2. What are the key differences between solving differential equations with Laplace transforms and other methods?
  3. Why is the term y13sin(3t)\frac{y_1}{3} \sin(3t) divided by 3 in this solution?
  4. Can Laplace transforms be used for non-homogeneous differential equations?
  5. What other real-world problems can be solved using Laplace transforms?

Tip:

Always check the initial conditions after solving the differential equation to ensure the solution satisfies them!

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Laplace Transform
Inverse Laplace Transform

Formulas

Laplace transform of y'(t): L(y'(t)) = sY(s) - y(0)
Laplace transform of y''(t): L(y''(t)) = s^2Y(s) - sy(0) - y'(0)
Inverse Laplace transform pairs for sine and cosine

Theorems

Linear Differential Equations
Laplace Transform Theorem

Suitable Grade Level

Undergraduate (Engineering or Applied Math)