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By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}-4 y=4 e^{2 t}, \quad y_{0}=0, y_{0}^{\prime}=1$$

Short Answer

Expert verified
The solution to the differential equation is \(y(t) = \frac{1}{2}e^{2t} - \frac{1}{2}e^{-2t} + t e^{2t}\).

Step by step solution

01

- Apply the Laplace Transform

Take the Laplace transform of both sides of the differential equation. Recall that the Laplace transform of a derivative is given by \(\text{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)\), \(\text{L}\{y(t)\} = Y(s)\), and \(\text{L}\{e^{2t}\} = \frac{1}{s-2}\).
02

- Substitute Initial Conditions

Substitute the initial conditions \(y(0) = 0\) and \(y'(0) = 1\) into the transformed equation. This gives: \((s^2Y(s) - sy(0) - y'(0)) - 4Y(s) = 4 \text{L}\{e^{2t}})\).
03

- Simplify the Transformed Equation

Simplify the equation using the initial conditions: \(s^2Y(s) - s \times 0 - 1 - 4Y(s) = 4 \times \frac{1}{s-2}\), which simplifies to \(s^2Y(s) - 1 - 4Y(s) = \frac{4}{s-2}\).
04

- Solve for Y(s)

Combine terms and solve for \(Y(s)\): \(Y(s)(s^2 - 4) = \frac{4}{s-2} + 1\). Rearrange to obtain: \(Y(s) = \frac{4}{(s-2)(s^2-4)} + \frac{1}{s^2-4}\).
05

- Partial Fraction Decomposition

Perform partial fraction decomposition on \(Y(s)\): \(\frac{4}{(s-2)(s^2-4)} = \frac{4}{(s-2)(s-2)(s+2)}\). Decompose into simpler fractions and solve for the constants.
06

- Inverse Laplace Transform

Apply the inverse Laplace transform to each term individually. Use known transforms such as \(L^{-1}\{ \frac{1}{s-a} \} = e^{at} \) and \( L^{-1}\{ \frac{1}{s^2-a^2} \} = \frac{\text{sinh}(at)}{a}\). Combine the inverse transforms to find \(y(t)\).

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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 equations involving derivatives of an unknown function. They can describe a wide range of phenomena such as movement, growth, and decay.
In this exercise, we deal with a second-order linear differential equation: \( y'' - 4y = 4e^{2t} \). Here,
  • \( y'' \) is the second derivative of \( y(t) \)
  • \( y \) is the original function.
Our goal is to find \( y(t) \), given certain initial conditions (we'll address those next).
Initial Conditions
Initial conditions are values given for the function and its derivatives at specific points. These are critical because they allow us to find a unique solution to the differential equation.
In this problem, the initial conditions are:\( y(0) = 0 \) and \( y'(0) = 1 \).
When we take the Laplace transform of a differential equation, these conditions help us simplify the transformed equation. Specifically, they will be used to replace terms involving \( y(0) \) and \( y'(0) \) as we see in Step 2, making our equation more solvable.
Laplace Transform Properties
The Laplace Transform is a powerful tool for solving differential equations. It changes functions from the time domain into the complex frequency domain. The main properties we use include:
  • \( \text{L}\big\{ y'(t) \big\} = sY(s) - y(0) \)
  • \( \text{L}\big\{ y''(t) \big\} = s^2Y(s) - sy(0) - y'(0) \)
  • \( \text{L}\big\{ e^{at} \big\} = \frac{1}{s-a} \)
These transformations allow us to convert our differential equation into an algebraic form that is easier to handle.When solving for \( Y(s) \), which represents our function in the frequency domain, we can then manipulate it algebraically to solve for \( y(t) \) in the time domain.
Inverse Laplace Transform
Once we have an algebraic expression for \( Y(s) \), we need to convert it back to the time domain using the Inverse Laplace Transform. This often involves partial fraction decomposition to simplify the expression.
Important transforms include:
  • \( L^{-1}\big\{\frac{1}{s-a}\big\} = e^{at} \)
  • \( L^{-1}\big\{\frac{1}{s^2-a^2}\big\} = \frac{\text{sinh}(at)}{a} \)
By applying these inverse transforms to each term of \( Y(s) \), we get our final solution in terms of \( y(t) \).

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Most popular questions from this chapter

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}-6 y^{\prime}+9 y=t e^{3 t}, \quad y_{0}=0, y_{0}^{\prime}=5$$

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}-5 y^{\prime}+6 y=e^{2 x}$$

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

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}+4 y^{\prime}+5 y=26 e^{3 x}$$

An object of mass \(m\) falls from rest under gravity subject to an air resistance proportional to its speed. Taking the \(y\) axis as positive down, show that the differential equation of motion is \(m(d v / d t)=m g-k v,\) where \(k\) is a positive constant. Find \(v\) as a function of \(t,\) and find the limiting value of \(v\) as \(t\) tends to infinity; this limit is called the terminal speed. Can you find the terminal speed directly from the differential equation without solving it? Hint: What is \(d v / d t\) after \(v\) has reached an essentially constant value? Consider the following specific examples of this problem. (a) A person drops from an airplane with a parachute. Find a reasonable value of \(k\) (b) In the Millikan oil drop experiment to measure the charge of an electron, tiny electrically charged drops of oil fall through air under gravity or rise under the combination of gravity and an electric field. Measurements can be made only after they have reached terminal speed. Find a formula for the time required for a drop starting at rest to reach 99\% of its terminal speed.

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