Question

In Exercises \(1-20\) solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ 4 y^{\prime \prime}-4 y^{\prime}+5 y=4 \sin t-4 \cos t+\delta(t-\pi / 2)-\delta(t-\pi), \quad y(0)=1, \quad y^{\prime}(0)=1 $$

Short answer

Question: Find the solution to the initial value problem with a second-order, nonhomogeneous linear differential equation given by: $$ 4y'' - 4y' + 5y = 4\sin{t} - 4\cos{t} + 12\delta(t - \frac{\pi}{2}) - 12\delta(t - \pi), $$ where \(\delta\) represents the Dirac delta function, and the initial conditions are \(y(0) = 1\) and \(y'(0) = 1\). Answer: The solution to the initial value problem is: $$ y(t) = 2e^{\frac{1}{2}t}\cos{\frac{\sqrt{3}}{2}t} + e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}t} -\sin{t} -\cos{t} -\frac{2\sqrt{3}}{3}e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi / 2)}+\frac{2\sqrt{3}}{3}e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi)}. $$

Step-by-step solution

Find the complementary function (yc)

First, we need to find the complementary function (yc) by solving the associated homogeneous differential equation: $$ 4y''-4y'+5y=0 $$ To do this, we'll assume a solution of the form \(y=e^{rt}\), where r is a constant. By plugging this into the homogeneous differential equation and dividing by \(e^{rt}\), we get a characteristic equation: $$ 4r^2 - 4r + 5 = 0 $$ By solving the quadratic equation, we find that \(r=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\). The complementary function, yc, is given by: $$ y_c (t) = C_1 e^{\frac{1}{2}t}\cos{\frac{\sqrt{3}}{2}t} +C_2 e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}t} $$

Determine a particular solution for the sinusoidal terms

Next, we will find a particular solution, called \(y_p(t)\), for the sinusoidal terms on the right-hand side of the given differential equation. To do this, we will assume the solution has the form: $$ y_p(t)=A\sin{t}+B\cos{t} $$ We'll differentiate this twice to find \(y_p'(t)\) and \(y_p''(t)\), and then we'll substitute these into the given differential equation to solve for A and B. $$ y_p'(t)=A\cos{t}-B\sin{t}, \quad y_p''(t)=-A\sin{t}-B\cos{t} $$ Substitute \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into the given differential equation to solve for A and B: $$ 4(-A\sin{t}-B\cos{t})-4(A\cos{t}-B\sin{t})+5(A\sin{t}+B\cos{t})=4\sin{t}-4\cos{t} $$ By equating coefficients for the sine and cosine terms, we can solve for A and B: $$ (5-4)A +(4)B =4,\quad(-4)A+(4-4)B=-4. $$ Solving this system, we find that \(A=-1\) and \(B=-1\). The particular solution is: $$ y_p(t)=-\sin{t}-\cos{t} $$

Find particular solutions for the delta functions

Now we will find particular solutions \(y_{p1}(t)\) and \(y_{p2}(t)\) for the two delta function terms: $$ y_{p1}(t)=C_3e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi / 2)}, \, \, y_{p2}(t)=C_4e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi)}. $$

Combine the solutions

Now let's combine the solutions from steps 1-3 to form the general solution: $$ y(t)=y_c(t)+y_p(t)+y_{p1}(t)+y_{p2}(t)\\ y(t)=C_1e^{\frac{1}{2}t}\cos{\frac{\sqrt{3}}{2}t} +C_2e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}t}-\sin{t}-\cos{t}\\ + C_3e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi / 2)}+C_4e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi)} $$

Apply the initial conditions

Now we'll apply the initial conditions \(y(0)=1\) and \(y'(0)=1\) to find the constants \(C_1\), \(C_2\), \(C_3\), and \(C_4\) in the general solution: $$ y(0)=C_1(1)+C_2(0)-1(0)-1(1)+C_3(0)+C_4(0)=1, $$ $$ y'(0)=\frac{1}{2}C_1(0)+C_2\frac{\sqrt{3}}{2}(1)-1(1)-1(0)+ C_3\left(\frac{\sqrt{3}}{2}\right)+C_4(0)=1. $$ Solving this system, we find that \(C_1=2\), \(C_2=1\), \(C_3=-\frac{2\sqrt{3}}{3}\), and \(C_4=\frac{2\sqrt{3}}{3}\). Thus, the solution for the initial value problem is: $$ y(t)=2e^{\frac{1}{2}t}\cos{\frac{\sqrt{3}}{2}t} +e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}t}-\sin{t}-\cos{t} - \frac{2\sqrt{3}}{3}e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi / 2)} + \frac{2\sqrt{3}}{3}e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi)} $$

Key concepts

Complementary Function

When working with differential equations, the complementary function is a crucial part of the solution. It's the solution to the homogeneous version of the differential equation. By solving this simpler equation, we look for functions that align with its conditions without considering independent parts, like external forces or particular solutions.
To find the complementary function, assume a solution in exponential form. This often involves a guess of the form \( y = e^{rt} \), where \( r \) is a constant, allowing us to derive a characteristic equation.

In our example, solving the characteristic equation \( 4r^2 - 4r + 5 = 0 \), we get complex roots. This indicates oscillatory behavior in solutions with the form:

• \( y_c (t) = C_1 e^{\frac{1}{2}t}\cos{\frac{\sqrt{3}}{2}t} + C_2 e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}t} \).

These terms represent the complementary function and are crucial for the general solution of the differential equation.

Particular Solution

A particular solution accounts for the non-homogeneous part of a differential equation. This component addresses terms that alter the solution beyond the homogeneous part, integrating the specific influences described by the non-homogeneous part.
To find the particular solution, we typically use an ansatz or educated guess based on the form of non-homogeneous terms.

• In our exercise, the guess \( y_p(t) = A\sin t + B\cos t \) was used for the sinusoidal terms.
• Upon substitution into the differential equation, coefficients of sine and cosine are equated to find \( A = -1 \) and \( B = -1 \).

This led us to the particular solution \( y_p(t) = -\sin t - \cos t \). By solving for these coefficients, this solution captures the unique influences of those non-homogeneous terms.

Delta Function Solution

Delta functions are unique as they represent sudden impacts or impulses occurring at specific times. In differential equations, they act as point sources that can change the outcomes in ways not captured by regularly continuous functions.
To find the response of a system to a delta function, solutions typically employ terms associated with the locations of impulses. Examples in our problem include:

• \( y_{p1}(t) = C_3 e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi / 2)} \)
• \( y_{p2}(t) = C_4 e^{\frac{1}{2}t}\sin{\frac{\sqrt{3}}{2}(t-\pi)} \)

These solutions consider how a system responds specifically around the impulse's timing. Calculation of such solutions requires integration and initial condition adjustments.

Homogeneous Differential Equation

Homogeneous differential equations are the backbone for solving more complex non-homogeneous problems. These are equations set to zero, thus examining inherent properties of the system without external influences.
Analyzing the homogeneous differential equation involves finding solutions that naturally satisfy it when the independent functions (e.g., forcing terms) are absent.

• The sample problem's homogeneous form is \( 4y'' - 4y' + 5y = 0 \).

Upon finding solutions for this equation, namely the complementary function, we use these to inform and construct the broader general solution. Solving homogeneous equations helps us understand the underlying dynamics and stability of a system under assumptions of ideal conditions.