Chapter 8: Problem 15
By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+9 y=\cos 3 t, \quad y_{0}=0, y_{0}^{\prime}=6$$
Short Answer
Expert verified
The solution is y(t) = 2t.
Step by step solution
01
Take the Laplace Transform of the Differential Equation
Use the property of Laplace transforms to convert the differential equation into an algebraic equation. The Laplace transform of the given differential equation is: \[ L\{y^{\prime\prime} + 9y\} = L\{\cos 3t\} \] Using the linearity of the Laplace transform, we get: \[ L\{y^{\prime\prime}\} + 9L\{y\} = L\{\cos 3t\} \]
02
Apply Laplace Transform Properties
Applying the Laplace transform properties: \[ L\{y^{\prime\prime}\} = s^2Y(s) - sy(0) - y^{\prime}(0) \] and \[ L\{\cos 3t\} = \frac{s}{s^2+9} \] Substituting the initial conditions, we have: \[ y(0) = 0 \] and \[ y^{\prime}(0) = 6 \] Thus, the transformed equation is: \[ s^2Y(s) - sy(0) - y^{\prime}(0) + 9Y(s) = \frac{s}{s^2 + 9} \] Which simplifies to: \[ s^2Y(s) - 0s - 6 + 9Y(s) = \frac{s}{s^2 + 9} \] \[ (s^2 + 9)Y(s) - 6 = \frac{s}{s^2 + 9} \]
03
Solve for Y(s)
Rearrange to solve for \(Y(s)\): \[ Y(s)(s^2 + 9) - 6 = \frac{s}{s^2 + 9} \] Multiply both sides by \(s^2 + 9\): \[ Y(s)(s^2 + 9)^2 - 6(s^2 + 9) = s \] Isolate \(Y(s)\): \[ Y(s)(s^4 + 18s^2 + 81) = s + 6(s^2 + 9) \] Expand and simplify: \[ Y(s)(s^4 + 18s^2 + 81) = s + 6s^2 + 54 \] \[ Y(s) = \frac{s + 6s^2 + 54}{s^4 + 18s^2 + 81} \]
04
Simplify and Perform Inverse Laplace Transform
Simplify and prepare for inversion: \[ Y(s) = \frac{s(s^2 + 9) + 6s^2 + 54}{(s^2 + 9)^2} \] \[ Y(s) = \frac{6s + 54}{((s^2 + 9)^2)} \] Perform the inverse Laplace transform: Using known transforms, the solution in the time domain is: \[ y(t) = 2t \]
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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 mathematical technique used to transform functions from the time domain to the frequency domain. This is particularly useful for solving linear differential equations.
It converts complex differential equations into simpler algebraic equations. The general formula for the Laplace transform of a function \( f(t) \) is:
\[ L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t \]
Here, \( s \) is a complex number frequency parameter, and \( L \{ f(t) \} \) represents the Laplace transform of the function \( f(t) \).
The key properties of Laplace transforms that make them useful include linearity, differentiation, and initial conditions. In the context of our exercise, it allows us to transform the given differential equation \( y'' + 9y = \cos(3t) \) into a more manageable algebraic equation.
In our solution, we first took the Laplace transform of both sides of the differential equation, giving us:
\[ L\{ y'' + 9y \} = L\{ \cos(3t) \} \]
It converts complex differential equations into simpler algebraic equations. The general formula for the Laplace transform of a function \( f(t) \) is:
\[ L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t \]
Here, \( s \) is a complex number frequency parameter, and \( L \{ f(t) \} \) represents the Laplace transform of the function \( f(t) \).
The key properties of Laplace transforms that make them useful include linearity, differentiation, and initial conditions. In the context of our exercise, it allows us to transform the given differential equation \( y'' + 9y = \cos(3t) \) into a more manageable algebraic equation.
In our solution, we first took the Laplace transform of both sides of the differential equation, giving us:
\[ L\{ y'' + 9y \} = L\{ \cos(3t) \} \]
Differential Equations
A differential equation is an equation involving a function and its derivatives. These equations describe how a particular quantity changes over time.
In the given exercise, we are dealing with a second-order linear differential equation with constant coefficients:
\[ y'' + 9y = \cos(3t) \]
This equation involves the second derivative \( y'' \) and the function \( y \) itself. To solve it using the Laplace transform, we turn this equation into an algebraic equation in terms of \( Y(s) \), the Laplace transform of \( y(t) \).
Once this transformation is done, we work with algebraic methods to solve for \( Y(s) \). Finally, we apply the inverse Laplace transform to obtain \( y(t) \), the original function of time.
In the given exercise, we are dealing with a second-order linear differential equation with constant coefficients:
\[ y'' + 9y = \cos(3t) \]
This equation involves the second derivative \( y'' \) and the function \( y \) itself. To solve it using the Laplace transform, we turn this equation into an algebraic equation in terms of \( Y(s) \), the Laplace transform of \( y(t) \).
Once this transformation is done, we work with algebraic methods to solve for \( Y(s) \). Finally, we apply the inverse Laplace transform to obtain \( y(t) \), the original function of time.
Initial Conditions
Initial conditions are the values of the function and its derivatives at the starting point (usually \( t = 0 \)). They provide the necessary information to uniquely determine the solution of a differential equation.
For our exercise, the initial conditions are given as:
\[ y(0) = 0 \]\[ y'(0) = 6 \]
These conditions allow us to find specific constants when solving differential equations. The Laplace transform of a derivative incorporates initial conditions directly. For example, the transform of the second derivative \( y'' \) becomes:
\[ L\{ y'' \} = s^2 Y(s) - sy(0) - y'(0) \]
Using our initial conditions, \( y(0) = 0 \) and \( y'(0) = 6 \), we can substitute to transform the equation properly and promise a unique and accurate solution.
For our exercise, the initial conditions are given as:
\[ y(0) = 0 \]\[ y'(0) = 6 \]
These conditions allow us to find specific constants when solving differential equations. The Laplace transform of a derivative incorporates initial conditions directly. For example, the transform of the second derivative \( y'' \) becomes:
\[ L\{ y'' \} = s^2 Y(s) - sy(0) - y'(0) \]
Using our initial conditions, \( y(0) = 0 \) and \( y'(0) = 6 \), we can substitute to transform the equation properly and promise a unique and accurate solution.
Inverse Laplace Transform
The inverse Laplace transform is the process of converting back from the frequency domain to the time domain.
After solving for \( Y(s) \) in the frequency domain, we need the inverse transform to find the actual solution \( y(t) \). The inverse Laplace transform of a function \( F(s) \) is given by:
\[ L^{-1} \{ F(s) \} = f(t) \]
In our solution, we reached the form of \( Y(s) \) as:
\[ Y(s) = \frac{s + 6s^2 + 54}{(s^2 + 9)^2} \]
To find \( y(t) \), we decomposed it into simpler fractions and used known inverse transforms. The resulting function from the inverse Laplace transform gave us the final solution:
\[ y(t) = 2t \]Understanding this process helps in finding solutions to differential equations where direct methods are cumbersome.
After solving for \( Y(s) \) in the frequency domain, we need the inverse transform to find the actual solution \( y(t) \). The inverse Laplace transform of a function \( F(s) \) is given by:
\[ L^{-1} \{ F(s) \} = f(t) \]
In our solution, we reached the form of \( Y(s) \) as:
\[ Y(s) = \frac{s + 6s^2 + 54}{(s^2 + 9)^2} \]
To find \( y(t) \), we decomposed it into simpler fractions and used known inverse transforms. The resulting function from the inverse Laplace transform gave us the final solution:
\[ y(t) = 2t \]Understanding this process helps in finding solutions to differential equations where direct methods are cumbersome.
