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Chapter 8: Problem 3

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+4 y^{\prime}+4 y=e^{-2 t}, \quad y_{0}=0, y_{0}^{\prime}=4$$

Short Answer

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

Step by step solution

01

Take the Laplace Transform of Both Sides

First, apply the Laplace transform to both sides of the given differential equation. Use the linearity of the Laplace transform and the given initial conditions. The given differential equation is: \( y'' + 4y' + 4y = e^{-2t} \)Applying the Laplace transform, we get: \( L\{y'' + 4y' + 4y\} = L\{e^{-2t}\} \)
02

Use Properties of Laplace Transforms

Utilize the properties of Laplace transforms for each term in the equation. Recall that \( L\{y''\} = s^2 Y(s) - sy(0) - y'(0) \), \( L\{y'\} = s Y(s) - y(0) \), and\( L\{y\} = Y(s) \). Also, use the initial conditions: \( y(0) = 0 \) and \( y'(0) = 4 \). Thus, we have: \( s^2 Y(s) - s \cdot 0 - 4 + 4(s Y(s) - 0) + 4 Y(s) = \frac{1}{s+2} \)
03

Simplify the Equation

Combine like terms of the transformed differential equation to solve for \(Y(s)\). \[ s^2 Y(s) + 4s Y(s) + 4 Y(s) - 4 = \frac{1}{s+2} \]This simplifies to:\[ (s^2 + 4s + 4) Y(s) - 4 = \frac{1}{s+2} \]Factoring, we get:\[ (s+2)^2 Y(s) - 4 = \frac{1}{s+2} \]
04

Solve for \(Y(s)\)

Rearrange the simplified equation to solve for \(Y(s)\): \[ (s+2)^2 Y(s) = \frac{1}{s+2} + 4 \]Divide both sides by \((s+2)^2\) to isolate \(Y(s)\):\[ Y(s) = \frac{1}{(s+2)^3} + \frac{4}{(s+2)^2} \]
05

Take the Inverse Laplace Transform

Find the inverse Laplace transform of each term individually to obtain \(y(t)\).Use the known Laplace inverse transforms:\(L^{-1} \{ \frac{1}{(s+2)^3} \} = \frac{t^2}{2} e^{-2t} \)And,\(L^{-1} \{ \frac{4}{(s+2)^2} \} = 4t e^{-2t} \).Thus,\( y(t) = \frac{t^2}{2} e^{-2t} + 4t e^{-2t} \).

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Key Concepts

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

differential equations
Differential equations are mathematical expressions that relate a function with its derivatives. In this exercise, the differential equation provided is: \( y'' + 4y' + 4y = e^{-2t} \). When solving differential equations using Laplace transforms, we convert the problem into an algebraic equation in the s-domain. This much simpler equation can be solved using basic algebraic methods.
initial conditions
Initial conditions are essential for solving differential equations because they give information about the state of the system at the start (t=0). Here, we have the initial conditions: \( y(0) = 0 \) and \( y'(0) = 4 \). These values are used when we transform the differential equation using Laplace transforms. Specifically, the values of y and its derivatives at t=0 are plugged into the transformed equations to solve for the unknowns in terms of Y(s).
inverse Laplace transform
The inverse Laplace transform is used to revert a function from the s-domain back to the original t-domain. After solving the transformed algebraic equation, we need to apply the inverse Laplace transform to find the original function. In this case, after isolating \( Y(s) \), we applied the known inverse transforms: \[ L^{-1} \left\{ \frac{1}{(s+2)^3} \right\} = \frac{t^2}{2} e^{-2t} \] and \[ L^{-1} \left\{ \frac{4}{(s+2)^2} \right\} = 4t e^{-2t} \]. Combining these results, we obtain the solution \( y(t) = \frac{t^2}{2} e^{-2t} + 4t e^{-2t} \).
properties of Laplace transform
The properties of Laplace transforms are leveraged to simplify the process of transforming differential equations. Some key properties used in the solution include:
  • Linearity: The Laplace transform of a sum is the sum of the Laplace transforms.
  • Differentiation: Laplace transforms convert differential operators into polynomial multipliers. For example, \( L\{y'\} \) becomes \( sY(s) - y(0) \) and \( L\{y''\} \) becomes \( s^2Y(s) - sy(0) - y'(0) \).
  • Time Shifting: This allows us to transform more complex functions involving exponentials and time shifts.

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Most popular questions from this chapter

Solve the algebraic equation $$D^{2}+(1+2 i) D+i-1=0$$ (note the complex coefficients) and observe that the roots are complex but not complex conjugates. Show that the method of solution of (5.6) (case of unequal roots) is correct here, and so find the general solution of $$y^{\prime \prime}+(1+2 i) y^{\prime}+(i-1) y=0$$

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$\frac{d^{2} r}{d t^{2}}-6 \frac{d r}{d t}+9 r=0$$

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+16 y=8 \cos 4 t, \quad y_{0}=0, y_{0}^{\prime}=8$$

Find a particular solution satisfying the given conditions. \(x y^{\prime}-y=x^{2}, \quad y=6\) when \(x=2\)

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