Chapter 8: Problem 13

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

Short Answer

Expert verified

The solution is: $ y(t) = 2 \cosh (2t) + 2 \sin (t) + 2 \cos (t) $.

Step by step solution

Take the Laplace transform of both sides of the differential equation

The given differential equation is: y'' + y = 5 \sinh (2t). Taking the Laplace transform of both sides, we use the linearity property and the Laplace transforms of derivatives: $ \mathcal{L}\{ y'' \} + \mathcal{L}\{ y \} = \mathcal{L}\{ 5 \sinh (2t) \} $.

Apply the Laplace transform properties

Using the Laplace transform definitions, we have: $ \mathcal{L}\{ y'' \} = s^2 Y(s) - s y(0) - y'(0) $ $ \mathcal{L}\{ y \} = Y(s) $, and $ \mathcal{L}\{ \sinh (2t) \} = \frac{2}{s^2 - 4} $. Substituting initial conditions $ y(0) = 0 $ and $ y'(0) = 2 $, we get: $ s^2 Y(s) - 2 + Y(s) = \frac{10}{s^2 - 4} $.

Simplify the equation

Now, simplify the equation to solve for $ Y(s) $: $ s^2 Y(s) + Y(s) = \frac{10}{s^2 - 4} + 2 $ Combine the terms involving $ Y(s) $: $ Y(s)(s^2 + 1) = \frac{10}{s^2 - 4} + 2 $ Isolate $ Y(s) $: $ Y(s) = \frac{10}{(s^2 - 4)(s^2 + 1)} + \frac{2}{s^2 + 1} $.

Perform partial fraction decomposition

Now, decompose $ \frac{10}{(s^2 - 4)(s^2 + 1)} $ using partial fractions: \replace correct fractions notation$ \frac{10}{(s^2 - 4)(s^2 + 1)} = \frac{As + B}{s^2 - 4} + \frac{Cs + D}{s^2 + 1} $. Matching coefficients gives: $ A = 0 $, $ B = 2 $, $ C = 2 $, $ D = 0 $. Thus, the partial fractions are: $ \frac{10}{(s^2 - 4)(s^2 + 1)} = \frac{2}{s^2 - 4} + \frac{2s}{s^2 + 1} $.

Combine the partial fractions

Combine the obtained partial fractions: $ Y(s) = \frac{2}{s^2 - 4} + \frac{2s}{s^2 + 1} + \frac{2}{s^2 + 1} $.

Take the inverse Laplace transform

Now, take the inverse Laplace transform using the linearity property: $ y(t) = 2 \cosh (2t) + 2 \sin (t) + 2 \cos (t) $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laplace Transform

The Laplace Transform is a powerful tool for solving differential equations. It is defined as: \ $ \mathcal{L}\{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) dt $. This transform changes functions of time into functions of a complex variable $s$. For example, to solve a differential equation, we first take the Laplace transform of both sides of the equation. This transforms the differential equation into an algebraic equation, which is often easier to solve. Once the algebraic equation is solved for $Y(s)$, we take the inverse Laplace Transform to return to the original time domain.

Differential Equations

Differential equations involve functions and their derivatives. They describe the rate of change of a quantity. In the equation we solved, $ y'' + y = 5 \sinh 2 t $, $y''$ represents the second derivative of $y$ with respect to time $t$. Solving differential equations often involves finding a function $y(t)$ that meets the criteria set by the equation. By using the Laplace transform, we can convert the differential equation into a simpler algebraic form.

Initial Conditions

Initial conditions are values given for the function and its derivatives at the start point (usually $t = 0$). They are essential for solving differential equations, as they allow us to find specific solutions. In our example, the initial conditions were $ y(0) = 0 $ and $ y'(0) = 2 $. When applying the Laplace Transform, these values substitute into the transformed equation to help solve for $Y(s)$.

Partial Fraction Decomposition

Partial Fraction Decomposition is a method used to break down complex fractions into simpler parts. This is particularly useful in inverse Laplace Transforms. For example, in our solution, we decomposed $ \frac{10}{(s^2 - 4)(s^2 + 1)} $ into simpler fractions: $ \frac{2}{s^2 - 4} + \frac{2s}{s^2 + 1} $. This step simplifies the inverse Laplace transform process.

Inverse Laplace Transform

The Inverse Laplace Transform converts a function from the $s$-domain back to the $t$-domain. In our example, after partial fraction decomposition and collecting terms, we applied the inverse Laplace transform to find: $ y(t) = 2 \cosh (2t) + 2 \sin (t) + 2 \cos (t) $. This function is the solution to the original differential equation in terms of time.