Question

Express the solution of the given initial value problem in terms of a convolution integral. \(y^{\mathrm{iv}}+5 y^{\prime \prime}+4 y=g(t) ; \quad y(0)=1, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0, \quad y^{\prime \prime \prime}(0)=0\)

Short answer

In this exercise, we start with a fourth-order differential equation given by \(y^{\mathrm{iv}}+5 y^{\prime \prime}+4 y=g(t)\). The impulse response \(h(t)\) is found by solving the homogeneous equation with initial conditions for an impulse, which results in \(h(t)=\cos(t)+ \frac{1}{2} \cos(2t)\). The solution to the original differential equation for any input function \(g(t)\) can be expressed as a convolution integral: \(y(t)=\int_{0}^{t} h(\tau)g(t-\tau) d\tau\).

Step-by-step solution

Write the differential equation in standard form

In this case, the given differential equation is already in standard form: \(y^{\mathrm{iv}}+5 y^{\prime \prime}+4 y=g(t)\)

Find the impulse response h(t)

The impulse response is the solution to the homogeneous equation with initial conditions corresponding to an impulse. The homogeneous equation is: \(y^{\mathrm{iv}}+5 y^{\prime \prime}+4 y=0\) To solve this, we need to find the characteristic equation and its roots: \(r^4+5r^2+4=0\) Factoring the above equation, we get \((r^2+1)(r^2+4)=0\) The roots are \(r_1=i, r_2=-i, r_3=2i, r_4=-2i\) Since we have complex roots, the general solution to the homogeneous differential equation is given by: \(y_h(t)=C_1 \cos(t)+C_2 \sin(t)+C_3 \cos(2t)+C_4 \sin(2t)\) Now, we apply the initial conditions for an impulse: \(y(0)=1, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0, \quad y^{\prime \prime \prime}(0)=0\) Using these initial conditions, we can solve for \(C_1, C_2, C_3,\) and \(C_4\). After solving, we get: \(h(t)=y_h(t)=\cos(t)+ \frac{1}{2} \cos(2t)\)

Use the convolution integral to find the solution for input g(t)

Now that we have the impulse response \(h(t)\), we can use the convolution integral to find the solution to the original differential equation given any input function \(g(t)\): \(y(t)=h(t) * g(t) = \int_{0}^{t} h(\tau)g(t-\tau) d\tau\) This convolution integral represents the solution to the initial value problem for any input function \(g(t)\). In this particular exercise, the important part to show is the impulse response and the expression of the solution as a convolution integral.

Key concepts

Impulse Response

When dealing with linear time-invariant systems, the impulse response is a critical concept. It's the system's output when subjected to an impulse input. In simpler terms, an impulse response helps determine how the system reacts to a sudden force or signal at a specific moment.

For differential equations, especially those involving initial value problems, the impulse response describes how the system evolves in response to a delta function. In the initial problem provided, the impulse response was found by solving the homogeneous differential equation. This requires setting up initial conditions synonymous with an 'impulse,' typically by using delta functions, which are simplistic yet mathematically complex spike-like inputs.

The impulse response found, \[h(t) = \cos(t) + \frac{1}{2}\cos(2t),\]effectively captures the behavior of the system's output over time.

Characteristic Equation

The characteristic equation plays a vital role when solving linear differential equations. It arises from substituting a trial solution into the differential equation and equating polynomial coefficients. For the listed problem, the homogeneous differential equation was manipulated through its highest derivative to form the characteristic polynomial:

\[r^4 + 5r^2 + 4 = 0.\]

This equation is solved for its roots, which reveal information about the nature of solutions. Here, factoring gives us two sets of complex conjugate roots: \(r = i, -i, 2i, -2i\). These roots were crucial for constructing the full solution to the homogeneous equation via trigonometric functions using Euler's formula.

By understanding the characteristic equation and its roots, it's possible to determine patterns in solutions, such as oscillatory behavior or exponential growth/decay, depending on the nature of the roots.

Initial Value Problem

An initial value problem (IVP) specifies a differential equation along with values at a starting point. Solving an IVP involves using these starting conditions to find a unique solution that passes through the specified point.

In our example, the differential equation is provided with initial conditions:

• \(y(0) = 1\)
• \(y'(0) = 0\)
• \(y''(0) = 0\)
• \(y'''(0) = 0\)

These values are crucial for determining the constants in the general solution of the homogeneous differential equation. They underpin the process of developing a particular solution aligned with the system's initial conditions.

Overall, mastering how to apply initial conditions allows one to find precise solutions tailored to specific problems, thus making the differential equation relevant to real-world applications.

Homogeneous Differential Equation

A homogeneous differential equation is one where all terms involve the unknown function or its derivatives. Importantly, there is no term that stands alone (i.e., no constant or external forcing function like \(g(t)\) in this context). Such equations are typically of the form \(L[y] = 0\), where \(L\) denotes an operator involving derivatives.

The problem's homogeneous component, \[y^{iv} + 5y'' + 4y = 0,\]is solved by finding a general solution using methods like the characteristic equation. Complex roots indicate oscillatory solutions, forming parts of trigonometric functions, as encountered in solving this equation.

Understanding homogeneous differential equations is a stepping stone in superimposing particular solutions or external force responses, leading to comprehensive solutions involving particular and homogeneous elements.