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By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+4 y^{\prime}+5 y=26 e^{3 t}, \quad y_{0}=1, y_{0}^{\prime}=5$$

Short Answer

Expert verified
y(t) = 26 e^{3t} * u(t) - e^{-2t}(\text{cos}(t) - 2\text{sin}(t)) + e^{3 t} * u(t)

Step by step solution

01

- Take the Laplace Transform of Both Sides

Apply the Laplace transform to each term of the differential equation. Using the properties of Laplace transforms, we get: \(\text{L}\{y''(t)\text}\) + 4\(\text{L}\{y'(t)\}\) + 5\(\text{L}\{y(t)\}\) = \(\text{L}\{26e^{3t}\}\).
02

- Apply Initial Conditions for Laplace Transform

Substitute the initial conditions into the transformed equation: \(s^2 Y(s) - sy(0) - y'(0) + 4(s Y(s) - y(0)) + 5Y(s) = \frac{26}{s-3}\).
03

- Simplify the Equation

Substitute \(y(0) = 1\) and \(y'(0) = 5\): \(s^2 Y(s) - s - 5 + 4(s Y(s) - 1) + 5Y(s) = \frac{26}{s-3}\).Combine like terms to get: \((s^2 + 4s + 5)Y(s) - s - 9 = \frac{26}{s-3}\).
04

- Solve for \(Y(s)\)

Isolate \(Y(s)\): \(Y(s) = \frac{26}{(s-3)(s^2 + 4s + 5)} + \frac{s + 9}{s^2 + 4s + 5}\).
05

- Perform Partial Fraction Decomposition

Separate the equation into partial fractions. Notice that the second term already has a form that can be inverted using known Laplace transforms: \(Y(s) = \frac{26}{(s-3)((s+2)^2+1)} + \frac{s + 2 + 2}{(s+2)^2 + 1}\).
06

- Take the Inverse Laplace Transform

Using known inverse Laplace transforms: \(y(t) = 26 e^{3t} * \text{u}(t) - e^{-2t} (\text{cos}(t) - 2\text{sin}(t)) + e^{3t} * \text{u}(t)\).

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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 integral transform used to convert differential equations into algebraic equations. This makes them easier to solve. It transforms a time-domain function, typically denoted as \( f(t) \), into a complex frequency-domain function, typically denoted as \( F(s) \). The basic definition is:
\[ \text{L}\{f(t)\} = \frac{ \int_{0}^{\text{∞}} e^{-st} f(t) dt } \].
Any function that satisfies certain conditions can be transformed using this method. Laplace Transforms are especially useful for linear differential equations with constant coefficients.
Differential Equations
Differential equations involve unknown functions and their derivatives. They are classified based on the type and order of the derivatives involved. In this exercise, we deal with a second-order linear differential equation given by:
\[ y^{\prime \prime} + 4y^{\prime} + 5y = 26e^{3t} \].
Solving differential equations typically requires methods like Laplace Transforms, especially when initial conditions are given.
Initial Conditions
Initial conditions are values given for the function and its derivatives at specific points, usually at \( t = 0 \). For the given problem, the initial conditions are:
- \( y(0) = 1 \)
- \( y^{\prime}(0) = 5 \)
These conditions are substituted into the Laplace-transformed equations to reflect the specific situation. This helps to uniquely determine the solution of the differential equation.
Partial Fraction Decomposition
Partial Fraction Decomposition is a technique used to break down complex rational expressions into simpler fractions. This helps in finding the inverse Laplace Transform. For our equation, the term \( Y(s) \) needs to be decomposed into simpler fractions:
\[ Y(s) = \frac{26}{(s-3)(s^2+4s+5)} + \frac{s+9}{s^2+4s+5} \].
This step transforms the problem into multiple simpler problems, making the inverse Laplace Transform easier to apply.
Inverse Laplace Transform
The Inverse Laplace Transform is used to convert a function back from the frequency domain to the time domain. After performing partial fraction decomposition, we find inverse transforms for each term separately. For example:
- \( \frac{26}{(s-3)((s+2)^2+1)} \) uses the known inverse transform for a shifted exponential and a shifted, damped sinusoidal.
Combining the results, we get:
\[ y(t) = 26e^{3t} - e^{-2t}(\cos(t) - 2\sin(t)) + e^{3t} \].
Each term corresponds to the individual inverse transforms of the decomposed fractions. This final step provides the solution in the original time domain.

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Most popular questions from this chapter

Solve Problem either by Laplace transforms and the convolution integral or by Green functions. $$y^{\prime \prime}+y=\sec ^{2} t$$

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. $$(y+2 x) d x-x d y=0$$

(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 \(T.\)

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$\left(x \cos y-e^{-\sin y}\right) d y+d x=0$$

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