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Chapter 8: Problem 20

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}-8 y^{\prime}+16 y=32 t, \quad y_{0}=1, y_{0}^{\prime}=2$$

Short Answer

Expert verified
The solution is \( y(t) = \frac{t^2}{2} e^{4t} + 9t e^{4t} - 4e^{4t} + 8t \).

Step by step solution

01

- Take the Laplace transform of both sides

First, take the Laplace transform of the differential equation. Recall that the Laplace transform of a derivative is: \[ \mathcal{L}\{ y' \}(s) = sY(s) - y_0 \] \[ \mathcal{L}\{ y'' \}(s) = s^2 Y(s) - s y_0 - y_0' \] Using these, the transform of the given equation is: \[ \mathcal{L}\{ y'' - 8y' + 16y \} = \mathcal{L}\{ 32t \} \] Substituting the transforms, we get: \[ (s^2 Y(s) - s y_0 - y_0') - 8 (s Y(s) - y_0) + 16 Y(s) = \frac{32}{s^2} \]
02

- Substitute initial conditions

Now substitute the given initial conditions, \( y_0 = 1 \) and \( y_0' = 2 \), into the equation: \[ (s^2 Y(s) - s \cdot 1 - 2) - 8 (s Y(s) - 1) + 16 Y(s) = \frac{32}{s^2} \] This simplifies to: \[ s^2 Y(s) - s - 2 - 8s Y(s) + 8 + 16 Y(s) = \frac{32}{s^2} \]
03

- Solve for \( Y(s) \)

Combine like terms to solve for \( Y(s) \): \[ s^2 Y(s) - 8s Y(s) + 16 Y(s) = s + 6 + \frac{32}{s^2} \] Factor out \( Y(s) \): \[ (s^2 - 8s + 16) Y(s) = s + 6 + \frac{32}{s^2} \] \[ (s - 4)^2 Y(s) = s + 6 + \frac{32}{s^2} \] Therefore, \[ Y(s) = \frac{s + 6 + \frac{32}{s^2}}{(s - 4)^2} \]
04

- Simplify the equation

Separate the terms for easier inverse Laplace transform: \[ Y(s) = \frac{s + 6}{(s - 4)^2} + \frac{32}{s^2 (s - 4)^2} \]
05

- Perform partial fraction decomposition

Finding coefficients and solving yields: \[ \frac{32}{s^2 (s - 4)^2} = \frac{8}{s^2} - \frac{4}{s} + \frac{8}{(s - 4)^2} \]
06

- Take the inverse Laplace transform

Using the inverse Laplace transform, translate each term back to the time domain: \[ y(t) = \mathcal{L}^{-1}\left\{ \frac{s + 6}{(s - 4)^2} \right\} + \mathcal{L}^{-1}\left\{ \frac{8}{s^2} \right\} - \mathcal{L}^{-1}\left\{ \frac{4}{s} \right\} + \mathcal{L}^{-1}\left\{ \frac{8}{(s - 4)^2} \right\} \] Apply the inverse transforms: \[ y(t) = \left(t + \frac{t^2}{2}\right) e^{4t} + 8t - 4e^{4t} + 8t e^{4t} \]
07

- Simplify the result

Combine like terms for the final solution: \[ y(t) = \left(t + \frac{t^2}{2} + 8t \right) e^{4t} - 4e^{4t} + 8t \] Transform it to: \[ y(t) = \frac{t^2}{2} e^{4t} + 9t e^{4t} - 4e^{4t} + 8t \] Hence the final solution is: \[ y(t) = \frac{t^2}{2} e^{4t} + 9t e^{4t} - 4e^{4t} + 8t \]

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Key Concepts

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

Differential Equations
Differential equations involve functions and their derivatives. These equations showcase how a function changes with respect to an independent variable, typically time. For example, in physics, they describe how velocity changes over time, given acceleration. Differential equations are categorized based on their order (first, second, etc.) and linearity. The given exercise deals with a second-order linear differential equation: \(y'' - 8y' + 16y = 32t\). Here, we have the second derivative \(y''\), the first derivative \(y'\), and the function itself \(y\). Solving differential equations is crucial for modeling phenomena across various fields like engineering, physics, and economics.
Initial Conditions
Initial conditions specify the values of the function and its derivatives at the start. These are essential in finding unique solutions. In our exercise, the initial conditions are \(y(0) = 1\) and \(y'(0) = 2\). They help in simplifying the transformed differential equation. When we take the Laplace transform of the second-order differential equation, we use these initial conditions for solving the resulting algebraic equation. Initial conditions often depict starting points, such as initial temperature, population, or velocity, making the solution uniquely determined.
Inverse Laplace Transform
The Inverse Laplace Transform is used to revert back from the frequency domain to the time domain. The process involves rules and tables for common transforms. In our exercise, after solving the algebraic equation for \(Y(s)\) in the Laplace domain, we need the inverse Laplace transform to find \(y(t)\). For instance, \(\mathcal{L}^{-1} \left\{ \frac{8}{s^2} \right\} = 8t\) and \(\mathcal{L}^{-1} \left\{ \frac{8}{(s - 4)^2} \right\} = 8te^{4t}\). Understanding these inverses is crucial since it brings the physical interpretation back from the mathematical abstraction.

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Most popular questions from this chapter

Suppose that the rate at which you work on a hot day is inversely proportional to the excess temperature above \(75^{\circ} .\) One day the temperature is rising steadily, and you start studying at 2 p.m. You cover 20 pages the first hour and 10 pages the second hour. At what time was the temperature \(75^{\circ} ?\)

(a) Consider a light beam traveling downward into the ocean. As the beam progresses, it is partially absorbed and its intensity decreases. The rate at which the intensity is decreasing with depth at any point is proportional to the intensity at that depth. The proportionality constant \(\mu\) is called the linear absorption coefficient. Show that if the intensity at the surface is \(I_{0},\) the intensity at a distance \(s\) below the surface is \(I=I_{0} e^{-\mu s} .\) The linear absorption coefficient for water is of the order of \(10^{-2} \mathrm{ft}^{-1}\) (the exact value depending on the wavelength of the light and the impurities in the water). For this value of \(\mu,\) find the intensity as a fraction of the surface intensity at a depth of \(1 \mathrm{ft}\), 50 ft, 500 ft, 1 mile. When the intensity of a light beam has been reduced to half its surface intensity \(\left(I=\frac{1}{2} I_{0}\right),\) the distance the light has penetrated into the absorbing substance is called the half-value thickness of the substance. Find the half-value thickness in terms of \(\mu .\) Find the half-value thickness for water for the value of \(\mu\) given above. (b) Note that the differential equation and its solution in this problem are mathematically the same as those in Example 1, although the physical problem and the terminology are different. In discussing radioactive decay, we call \(\lambda\) the decay constant, and we define the half-life \(T\) of a radioactive substance as the time when \(N=\frac{1}{2} N_{0}\) (compare half-value thickness). Find the relation between \(\lambda\) and \(T.\)

Consider the differential equation \((D-a)(D-b) y=P_{n}(x),\) where \(P_{n}(x)\) is a polynomial of degree \(n\). Show that a particular solution of this equation is given by (6.24) with \(c=0 ;\) that is, \(y_{p}\) is \(\left\\{\begin{array}{l}\text { a polynomial } Q_{n}(x) \text { of degree } n \text { if } a \text { and } b \text { are both different from zero; } \\ x Q_{n}(x) \text { if } a \neq 0, \text { but } b=0 \\ x^{2} Q_{n}(x) \quad \text { if } a=b=0\end{array}\right.\) Hint: To show that \(Q_{n}(x)=\sum a_{n} x^{n}\) is a solution of the differential equation for a given \(P_{n}=\sum b_{n} x^{n},\) you have only to show that the coefficients \(a_{n}\) can be found so that \((D-a)(D-b) Q_{n}(x) \equiv P_{n}(x) .\) Equate coefficients of \(x^{n}, x^{n-1}, \cdots,\) to see that this is always possible if \(a \neq b\). For \(b=0,\) the differential equation becomes \((D-a) D y=P_{n} ;\) what is \(D y\) if \(y=x Q_{n} ?\) Similarly, consider \(D^{2} y\) if \(y=x^{2} Q_{n}\).

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}-4 y^{\prime}+4 y=6 e^{2 x}$$

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$u(1-v) d v+v^{2}(1-u) d u=0$$
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