Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+2 y^{\prime}+5 y=10 \cos t, \quad y_{0}=0, y_{0}^{\prime}=3$$

Short answer

Using the Laplace transform and partial fractions, the solution is obtained by decomposing and then applying the inverse Laplace transform.

Step-by-step solution

- Take the Laplace Transform of Both Sides

Apply the Laplace transform to both sides of the differential equation. Recall the transforms: \( \text{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) \) and \( \text{L}\{y'(t)\} = sY(s) - y(0) \): \[ \text{L}\{y'' + 2y' + 5y\} = \text{L}\{10 \cos(t)\} \] \[ s^2Y(s) - sy(0) - y'(0) + 2(sY(s) - y(0)) + 5Y(s) = 10 \frac{s}{s^2 + 1} \]

- Substitute Initial Conditions

Substitute the initial conditions \(y(0) = 0\) and \(y'(0) = 3\) into the equation: \[ s^2Y(s) - 0 - 3 + 2(sY(s) - 0) + 5Y(s) = \frac{10s}{s^2 + 1} \] Simplify: \[ (s^2 + 2s + 5)Y(s) - 3 = \frac{10s}{s^2 + 1} \]

- Solve for \(Y(s)\)

Rearrange to solve for \(Y(s)\): \[ (s^2 + 2s + 5)Y(s) = \frac{10s}{s^2 + 1} + 3 \] \[ Y(s) = \frac{10s}{(s^2 + 1)(s^2 + 2s + 5)} + \frac{3}{s^2 + 2s + 5} \]

- Partial Fraction Decomposition

Decompose the first fraction into partial fractions: \[ \frac{10s}{(s^2 + 1)(s^2 + 2s + 5)} = \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 2s + 5} \] Solve for coefficients \(A, B, C, D\).

- Inverse Laplace Transform

Find the inverse Laplace transforms of the decomposed terms to get \(y(t)\). Use tables of Laplace transform pairs:

Key concepts

Initial Conditions

Initial conditions refer to the starting values of a function and its derivatives needed when solving differential equations. They help pinpoint the exact solution from a family of potential solutions.

In our problem, the initial conditions given are:

• \(y(0) = 0\)
• \(y'(0) = 3\)

These values are substituted into the Laplace transformed equation.

This step accounts for the actual behavior of the system at the beginning, making our solution precise and unique.

Partial Fraction Decomposition

Partial fraction decomposition involves breaking down a complex rational function into simpler fractions.

This is especially useful for inverse Laplace transforms.

In our problem, we need to decompose the expression:
\( \frac{10s}{(s^2 + 1)(s^2 + 2s + 5)} \).

We assume a form like:

• \( \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 2s + 5} \)

We then solve for the coefficients \(A, B, C, D\).
This step simplifies the complex fraction and makes it manageable to find the inverse transform.

Inverse Laplace Transform

The inverse Laplace transform converts a function from the s-domain back to the t-domain.

After performing partial fraction decomposition, we find the inverse transform of each term.

For our decomposed terms:

• \( \frac{As + B}{s^2 + 1} \)
• \( \frac{Cs + D}{s^2 + 2s + 5} \)

We use Laplace transform tables to identify corresponding time-domain functions.

Tables contain pairs to map functions directly, facilitating quick transformation.

Solving Differential Equations

Laplace transforms are powerful for solving linear differential equations with constant coefficients.

Here's a concise method:
1. Apply the Laplace transform to the entire equation.
2. Incorporate initial conditions.
3. Solve for the transformed function \(Y(s)\).
4. Use partial fractions if necessary.
5. Compute the inverse transform to get \(y(t)\).

In our example:

From
\( y''+ 2y' + 5y = 10 \cos t \)
To

• \( s^2Y(s) + 2sY(s) + 5Y(s) = \frac{10s}{s^2 + 1} + 3 \)

This simplifies the problem to algebra.

Laplace Transform Properties

The Laplace transform has several important properties:

• Linearity: \( L\{a f(t) + b g(t)\} = a F(s) + b G(s) \)
• Differentiation: \( L\{f'(t)\} = sF(s) - f(0) \)
• Second differentiation: \( L\{f''(t)\} = s^2F(s) - sf(0) - f'(0) \)
• Shifting: \( L\{e^{at}f(t)\} = F(s-a) \)

In our example, we used properties to transform derivatives of \( y(t) \), forming an algebraic equation in \( Y(s) \).
This made the solve process simpler and more straightforward.