Question

Find the solution of the given initial value problem and draw its graph. \(y^{\prime \prime}+2 y^{\prime}+2 y=\delta(t-\pi) ; \quad y(0)=1, \quad y^{\prime}(0)=0\)

Short answer

Question: Solve the second-order linear differential equation with initial values and sketch the solution: \(y'' + 2y' + 2y = \delta(t-\pi)\) given \(y(0) = 1\) and \(y'(0) = 0\). Answer: The solution to the given differential equation is given by \(y(t) = e^{-t}(A\cos(t) + B\sin(t)) + (A-B)t + (1-A)\) with the initial conditions applied. The graph of the solution has a damped oscillatory behavior and an upward or downward trend depending on the value of (A-B). There will be a vertical dashed line at \(t = \pi\) to represent the discontinuity caused by the delta forcing term.

Step-by-step solution

Find the Complementary Solution

To find the complementary solution, we first need to solve \(y'' + 2y' + 2y = 0\). This is a second-order linear homogeneous differential equation. Let's assume the solution is given by \(y_c = e^{rt}\) and substitute this into the equation: \(r^2 e^{rt} + 2re^{rt} + 2e^{rt} = 0\) Since \(e^{rt}\) is nonzero, we can divide by it: \(r^2 + 2r + 2 = 0\) This quadratic equation has complex roots, which are given by \(r = -1 \pm i\). Hence, the complementary solution is in the form of: \(y_c(t) = e^{-t}(A\cos(t) + B\sin(t))\)

Find a Particular Solution

To find a particular solution, let's integrate the given equation. \(\int(y''+2y'+2y)dt = \int \delta(t-\pi) dt\) Applying integration to both sides, we get: \(y'+2y+C_1 = \int \delta(t-\pi) dt\) Integrating once more, we get: \(y+C_1t+C_2 = \int \left[\int \delta(t-\pi) dt\right] dt\) Since the integral of \(\delta(t-\pi) dt\) is a unit step function: \(y(t) = y_c(t) + \int \left[\int \delta(t-\pi) dt\right] dt\) This represents the solution of the given initial value problem.

Apply the Initial Conditions

Now, apply the initial conditions \(y(0) = 1\) and \(y'(0) = 0\) to find the constants \(A, B, C_1\), and \(C_2\). Plugging in the initial conditions, we get: \(1 = e^{0}(A\cos(0) + B\sin(0)) + C_2 = A+C_2 \Rightarrow C_2 = 1-A\) \(0 = -e^{0}(A\cos(0) + B\sin(0)) + e^{0}(B\cos(0) - A\sin(0)) + C_1 = -A+B+C_1 \Rightarrow C_1 = A-B\) Thus the particular solution is: \(y(t) = e^{-t}(A\cos(t) + B\sin(t)) + (A-B)t + (1-A)\)

Sketch the Graph

To sketch the graph of the given solution, plot the function: \(y(t) = e^{-t}(A\cos(t) + B\sin(t)) + (A-B)t + (1-A)\) Keep in mind that the function has damped oscillatory behavior due to the first term and an upward/downward trend depending on the value of (A-B). Observe that the solution has discontinuities at \(t = \pi\) because of the delta forcing term. On the graph, represent these discontinuities with a vertical dashed line at \(t = \pi\).

Key concepts

Linear Homogeneous Differential Equation

Understanding a linear homogeneous differential equation is fundamental to solving many problems in mathematics and physics. These are equations of the form
\[ a_n(t)y^{(n)} + a_{n-1}(t)y^{(n-1)} + \ ... + a_1(t)y' + a_0(t)y = 0 \]
where a's represent the coefficients and can be functions of t, and n indicates the order of the derivative. The key characteristic of a homogeneous equation is that the right-hand side is zero. This implies the property of linearity and superposition, where the sum of any two solutions is also a solution.
In our exercise example,
\[ y'' + 2y' + 2y = 0 \]
is a second-order linear homogeneous differential equation where the coefficients are constant. This particular form is characteristic for equations with exponential solutions involving complex numbers, which often translate to oscillatory motion in physical systems. In such cases, we use the ansatz,
\[ y_c = e^{rt} \]
to find the roots of the characteristic equation, leading us towards the complementary solution, an essential part of any differential equation's overall solution.

Complementary Solution

The complementary solution, denoted as y_c(t) in differential equations, is crucial for understanding the behavior of homogeneous equations. It represents the general solution to the homogeneous part of the differential equation without considering any external forces or inputs (like the delta function in the given exercise).
The complementary solution takes into account initial conditions and is often expressed in terms of constants. For example, in the presence of complex roots r = -1 \b1 i, as seen in our example, the complementary solution emerges as a combination of exponential, sine, and cosine functions:
\[ y_c(t) = e^{-t}(A\text{cos}(t) + B\text{sin}(t)) \]
These constants A and B will later be determined using initial conditions. The complementary solution is paramount as it indicates the system's natural response, such as a damped oscillation pattern that occurs without external influence.

Particular Solution

In contrast to the complementary solution, the particular solution addresses the non-homogeneous part of a differential equation, often representing an external force or an input. In the context of our initial value problem, we're looking for a solution y_p(t) that satisfies the entire equation inclusive of the delta function on the right side, not just the homogeneous portion.
For the equation given in our exercise,
\[ y'' + 2y' + 2y = \b\ta(t-\b\toi) \]
we need to find a specific y_p(t) that, when added to the complementary solution y_c(t), forms the general solution that also fulfills the external conditions imposed by \b\ta(t-\b\toi). This particular solution depends on the form of the non-homogeneous term. For example, in the step function or impulse cases like the delta function, specific techniques such as integration or Laplace transform are often employed.
In our step-by-step solution, we integrate to accommodate the delta function's effect and then use the given initial conditions to solve for any constants, thereby obtaining a particular solution that perfectly fits the scenario presented in the problem statement.