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Mobile App Chapter 8: Problem 47
Solve Problem either by Laplace transforms and the convolution integral or by Green functions. $$y^{\prime \prime}+y=\sec ^{2} t$$
Short Answer
Expert verified
The solution is \( y(t) = t \cos t - \frac{1}{2}t \sin t \).
Step by step solution
01
Identify the Differential Equation
The given differential equation is: \ \[ y'' + y = \sec^{2} t \]
02
Laplace Transform of Both Sides
Take the Laplace Transform of both sides of the differential equation. Use the property that \(L\{y''\} = s^2 Y(s) - sy(0) - y'(0)\) and \(L\{y\} = Y(s)\). Assume zero initial conditions: \ \[ L\{y'' + y\} = L\{\sec^{2} t \} \] \ This transforms to: \ \[ (s^2 + 1)Y(s) = L\{\sec^{2} t \} \]
03
Finding Laplace Transform of Right Side
Find the Laplace Transform of \(\sec^{2} t\). This can be done using a known table of transforms: \ \[ L\{\sec^{2} t \} = \frac{s}{s^2 + 1} - \frac{s^3}{s^2 + 1} \]
04
Solve for Y(s)
Substitute back into the transformed equation and solve for \(Y(s)\): \ \[ (s^2 + 1)Y(s) = \frac{s}{s^2 + 1} - \frac{s^3}{s^2 + 1} \] \ \[ Y(s) = \frac{s}{(s^2+1)^2} - \frac{s^3}{(s^2+1)^2} \]
05
Inverse Laplace Transform
Take the inverse Laplace transform of \(Y(s)\) to get \(y(t)\): \ \[ \mathcal{L}^{-1} \{ Y(s) \} = \mathcal{L}^{-1} \left( \frac{s}{(s^2+1)^2} \right) - \mathcal{L}^{-1} \left( \frac{s^3}{(s^2+1)^2} \right) \] \ Using known Laplace transforms: \ \[ y(t) = t \cos t - \frac{1}{2}t \sin t \]
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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 that relate a function to its derivatives. They describe processes where the rate of change of a quantity is related to the quantity itself. In this exercise, we are dealing with a second-order differential equation:
\[ y'' + y = \sec^{2} t \]
The second-order means it involves the second derivative of the unknown function \( y \) with respect to \( t \). Differential equations are used to model real-world phenomena like motion, heat, and wave propagation. Solving them typically involves finding a function \( y(t) \) that satisfies the equation for all permissible values of \( t \).
\[ y'' + y = \sec^{2} t \]
The second-order means it involves the second derivative of the unknown function \( y \) with respect to \( t \). Differential equations are used to model real-world phenomena like motion, heat, and wave propagation. Solving them typically involves finding a function \( y(t) \) that satisfies the equation for all permissible values of \( t \).
Laplace Transform
The Laplace transform is a powerful integral transform used to convert differential equations into algebraic equations, which are easier to manipulate. It transforms a function of time (\( t \)) into a function of a complex variable (\( s \)). For example, the Laplace transform of a function \( f(t) \) is given by:
\[ L\{f\}(s) = \int_{0}^{\infty} e^{-st}f(t) \,dt \]
Applying the Laplace transform to both sides of our differential equation and using the properties of transforms, we turn our original differential equation into an algebraic equation in terms of \( Y(s) \), which is the Laplace transform of \( y(t) \).
\[ L\{f\}(s) = \int_{0}^{\infty} e^{-st}f(t) \,dt \]
Applying the Laplace transform to both sides of our differential equation and using the properties of transforms, we turn our original differential equation into an algebraic equation in terms of \( Y(s) \), which is the Laplace transform of \( y(t) \).
Inverse Laplace Transform
The inverse Laplace transform is used to convert back from the \( s \)-domain to the time domain (\( t \)). If \( F(s) \) is the Laplace transform of \( f(t) \), the inverse Laplace transform is represented as:
\[ \mathcal{L}^{-1}\{F(s)\} = f(t) \]
In our problem, after transforming the given differential equation and solving for \( Y(s) \), we applied the inverse Laplace transform to find \( y(t) \). Using known inverse transforms from a table, we obtained:
\[ y(t) = t \cos t - \frac{1}{2}t \sin t \]
\[ \mathcal{L}^{-1}\{F(s)\} = f(t) \]
In our problem, after transforming the given differential equation and solving for \( Y(s) \), we applied the inverse Laplace transform to find \( y(t) \). Using known inverse transforms from a table, we obtained:
\[ y(t) = t \cos t - \frac{1}{2}t \sin t \]
Convolution Integral
The convolution integral is a method used to find the inverse Laplace transform of a product of two Laplace transforms. It is defined as:
\[ (f * g)(t) = \int_{0}^{t} f(\tau)g(t-\tau) \, d\tau \]
In our exercise, we solved for \( Y(s) \) and used known transforms to directly find the inverse without needing the convolution integral. However, in more complex problems where we have products of transforms that are not easily invertible, the convolution integral becomes essential. It's particularly useful when solving linear time-invariant systems.
\[ (f * g)(t) = \int_{0}^{t} f(\tau)g(t-\tau) \, d\tau \]
In our exercise, we solved for \( Y(s) \) and used known transforms to directly find the inverse without needing the convolution integral. However, in more complex problems where we have products of transforms that are not easily invertible, the convolution integral becomes essential. It's particularly useful when solving linear time-invariant systems.
Green's Functions
Green's functions are used to solve inhomogeneous differential equations. They provide a way to express the solution in terms of the original differential operator and a given forcing function. The general form is:
\[ y(t) = \int_{0}^{ \infty } G(t, \tau)f(\tau) \, d\tau \]
Where \( G(t, \tau) \) is the Green's function associated with the differential operator. In this particular exercise, we could have used Green's functions to find the solution, but instead, we used the Laplace transform method. Both methods are valid and often the choice depends on the problem at hand and the initial conditions provided.
\[ y(t) = \int_{0}^{ \infty } G(t, \tau)f(\tau) \, d\tau \]
Where \( G(t, \tau) \) is the Green's function associated with the differential operator. In this particular exercise, we could have used Green's functions to find the solution, but instead, we used the Laplace transform method. Both methods are valid and often the choice depends on the problem at hand and the initial conditions provided.
