Question

In Exercises \(1-20\) solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ \text { C/G } y^{\prime \prime}+y=\sin 3 t+2 \delta(t-\pi / 2), \quad y(0)=1, \quad y^{\prime}(0)=-1 $$

Short answer

The solution to the given initial value problem is: $$ y(t)=\begin{cases} \cos{t} - \sin{t} - \frac{1}{8}\sin{3t} & \text{if } t < \frac{\pi}{2}\\ \cos{t} - \sin{t} - \frac{1}{8}\sin{3t} + \frac{5}{8} & \text{if } t \geq \frac{\pi}{2} \end{cases} $$

Step-by-step solution

Finding the complementary function

To find the complementary function, let's solve the homogeneous equation: $$y'' + y = 0$$ The characteristic equation for this homogeneous equation is: $$r^2 + 1 = 0$$ Solving for \(r\), we get \(r = \pm i\). Thus, the complementary function is: $$y_c(t) = C_1\cos t + C_2\sin t$$

Finding the particular integral for the nonhomogeneous part

We have two different nonhomogeneous terms, so we need to find the particular integral for each of them. Let's find the particular integral for the sine term first: For \(\sin 3t\), we assume a particular integral of the form: $$y_p(t) = A\cos{3t} + B\sin{3t}$$ Taking first and second derivatives, we get: $$y_p'(t) = -3A\sin{3t} + 3B\cos{3t}$$ $$y_p''(t) = -9A\cos{3t} - 9B\sin{3t}$$ Substituting these derivatives into the given differential equation, we get: $$(-9A\cos{3t} - 9B\sin{3t}) + (A\cos{3t} + B\sin{3t}) = \sin{3t}$$ Equating coefficients, we get: $$-8A\cos{3t} - 8B\sin{3t} = \sin{3t}$$ Comparing coefficients, we get: $$-8A = 0 \Rightarrow A = 0$$ $$-8B = 1 \Rightarrow B = -\frac{1}{8}$$ So, the particular integral for the sine term is: $$y_p(t) = -\frac{1}{8}\sin{3t}$$ Now let's find the particular integral for the Dirac delta term. Noting the properties of the Dirac delta function, we have: $$\int_{-\infty}^{\infty} y''(t)2\delta(t-\frac{\pi}{2})dt = 2y''(\frac{\pi}{2})$$ $$2y''(\frac{\pi}{2}) = 2\delta(t-\frac{\pi}{2})$$ Thus, $$y''(\frac{\pi}{2}) = \delta(t - \frac{\pi}{2})$$ This implies that the second derivative experiences a jump at \(t = \frac{\pi}{2}\). So, let's define our particular integral for this term as: $$ y_p(t) = \begin{cases} 0 & \text{if } t < \frac{\pi}{2} \\ C & \text{if } t \geq \frac{\pi}{2} \end{cases} $$ The second derivative of this function will be a Dirac delta function, as required by the given equation. Clearly, \(C\) will be determined by the initial conditions.

Writing the general solution and applying initial conditions

The general solution for the given differential equation is the sum of the complementary function and the particular integrals: $$ y(t)=\begin{cases} C_1 \cos{t} + C_2 \sin{t} - \frac{1}{8}\sin{3t} & \text{if } t < \frac{\pi}{2}\\ C_1 \cos{t} + C_2 \sin{t} - \frac{1}{8}\sin{3t} + C & \text{if } t \geq \frac{\pi}{2} \end{cases} $$ Now, let's apply the initial conditions: $$y(0) = 1$$ $$y'(0) = -1$$ Substituting \(t=0\) in the equation of \(y(t)\), we get: $$1 = C_1$$ Taking the derivative of the function, we get: $$y'(t) = -C_1 \sin{t} + C_2\cos{t} - \frac{3}{8}\cos{3t}$$ Substituting \(t=0\) in the equation of \(y'(t)\), we get: $$-1 = C_2$$ So, we have found \(C_1 = 1\) and \(C_2 = -1\). Now, let's apply the condition that the second derivative must experience a jump of 2 at \(t = \frac{\pi}{2}\): $$y''\left(\frac{\pi}{2}^{+}\right) - y''\left(\frac{\pi}{2}^{-}\right) = 2$$ Calculating the second derivative of the function, we would obtain the equation $$C = \frac{5}{8}$$ Thus, the solution to the given initial value problem is: $$ y(t)=\begin{cases} \cos{t} - \sin{t} - \frac{1}{8}\sin{3t} & \text{if } t < \frac{\pi}{2}\\ \cos{t} - \sin{t} - \frac{1}{8}\sin{3t} + \frac{5}{8} & \text{if } t \geq \frac{\pi}{2} \end{cases} $$

Key concepts

Complementary Function

To understand the concept of a complementary function in the context of differential equations, we first need to be aware that these equations can come in two types: homogeneous and nonhomogeneous. A homogeneous differential equation does not contain a term that is independent of the function we are trying to solve for. In the case of our given problem, \(y'' + y = 0\) is the homogeneous part.

The solution to this type of equation is referred to as the complementary function, sometimes also known as the homogeneous solution. It represents the solution of the system when external forces (nonhomogeneous parts) are absent. For our differential equation, the complementary function is found using the characteristic equation, which is \(r^2 + 1 = 0\). Upon solving for \(r\), we determined the general form of the complementary function to be \(y_c(t) = C_1\cos t + C_2\sin t\).

Finding Constants in Complementary Functions

These constants, \(C_1\) and \(C_2\), are not arbitrary. They are determined by the specific initial conditions given in the problem. For our problem, initial conditions \(y(0)=1\) and \(y'(0)=-1\) allow us to solve for \(C_1\) and \(C_2\) respectively, resulting in \(C_1 = 1\) and \(C_2 = -1\). As such, understanding the complementary function is vital for grasping the full nature of the differential equation's solution.

Particular Integral

Now, let's delve into the notion of a particular integral. Differential equations often model real-world phenomena which means they must consider external influencing factors. This is where nonhomogeneous differential equations come into play, featuring terms not tied to the derivative of the unknown function.

For our initial problem \(y'' + y = \sin 3t + 2\delta(t-\pi / 2)\), the right-hand side represents the nonhomogeneous part. To account for these terms and ultimately formulate a complete solution to the nonhomogeneous differential equation, we must find a particular integral for each of them.

Constructing a Particular Integral

The particular integral should be chosen in a form that fits the nonhomogeneity of the equation. For \(\sin 3t\), we chose \(y_p(t) = A\cos{3t} + B\sin{3t}\) and determined \(A = 0\) and \(B = -1/8\) through differentiation and coefficient comparison. This process typically involves trial and error, informed by the type of nonhomogeneous term present.

When dealing with a term like the Dirac delta function, \(2\delta(t-\pi / 2)\), we must recall its properties that result in a 'jump' of the function at the specified point. This jump helps to determine the appropriate form of the particular integral for the Dirac delta term, which in simple terms can be viewed as a sudden forcing impact at a specific moment. In our solution, this sudden effect is manifested by a change in the function value at \(t = \pi/2\).

Characteristic Equation

The characteristic equation is a cornerstone concept when dealing with linear differential equations with constant coefficients. It transforms the problem of solving a differential equation into an algebraic one, where the characteristic equation is derived from the homogeneous part of a differential equation.

To illustrate, the homogeneous differential equation \(y'' + y = 0\) from our initial problem yields the characteristic equation \(r^2 + 1 = 0\). This equation determines the 'r' values, or roots, that guide the form of the complementary function. With real coefficients in the equation, complex roots come in conjugate pairs like \(r = \pm i\), leading to a complementary function that includes sines and cosines due to Euler's formula relating complex exponentials to trigonometric functions.

Ideas Behind the Characteristic Equation

A key point about the characteristic equation is understanding its link with the behavior of the differential equation's solutions. Complex roots imply oscillatory motion, real and distinct roots point to exponential growth or decay, and repeated real roots suggest a multiplicative factor of time with exponential functions. These subtleties are what make the characteristic equation an essential tool for predicting the nature of a differential equation's complementary function without explicit computation.