Problem 501 - Advanced Calculus

Question

Solve the initial value problem $$ \begin{aligned} &y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=H(t) \\ &y(0)=y^{\prime}(0)=0 \end{aligned} $$ where $$ \mathrm{H}(\mathrm{t})=1, \quad 0 \leq \mathrm{t}<\pi $$ $$ \text { and }=0, \quad t \geq \pi, $$ as shown in the accompanying graph.

Step-by-Step Solution

Answer

The final solution to the given initial value problem is \(y(t) = -\frac{1}{5}e^{-t}\cos(2t) + \frac{1}{10}e^{-t}\sin(2t) + \frac{1}{5}\).

Step 1: Solve the homogeneous differential equation

To find the complementary function, solve the homogeneous form of the given equation: \[ y^{\prime \prime}(t) + 2y^{\prime}(t) + 5y(t) = 0 \] The characteristic equation for this homogeneous equation is: \[ r^2 + 2r + 5 = 0 \] To solve for roots, use the quadratic formula given by \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 2\), and \(c = 5\). We have: \[ r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \] Since the discriminant \(b^2 - 4ac\) is negative, the roots will be complex conjugates.

Step 2: Complementary function

With complex conjugate roots, the complementary function can be expressed as: \[ y_c(t) = c_1 e^{at} \cos(bt) + c_2 e^{at} \sin(bt) \] where \(r = a \pm bi\) From Step 1, we have the characteristic equation's roots as follows: \[ r = -1 \pm 2i \] So, \(a = -1\) and \(b = 2\). Therefore, the complementary function is: \[ y_c(t) = c_1 e^{-t} \cos(2t) + c_2 e^{-t} \sin(2t) \]

Step 3: Particular solution

For the non-homogeneous equation, imagine a distinctive function as follows: \[ y_p(t) = A \] where A is a constant Using the given function H(t) in the range \(0 \le t < \pi\), we have: \[ y^{\prime \prime}(t) + 2y^{\prime}(t) + 5y(t) = 1 \] Since we chose a constant function for the particular solution, both its first and second derivatives are zero. So, the non-homogeneous equation becomes: \[ 5A = 1 \] Solving for A, we'll get the particular solution: \[ y_p(t) = \frac{1}{5} \]

Step 4: General solution

The general solution is the sum of the complementary function and the particular solution: \[ y(t) = y_c(t) + y_p(t) = c_1 e^{-t} \cos(2t) + c_2 e^{-t} \sin(2t) + \frac{1}{5} \]

Step 5: Apply initial conditions

We're given the initial conditions \(y(0) = y^{\prime}(0) = 0\). Let's apply them to our general solution. For \(y(0) = 0\): \[ 0 = c_1 e^{-0} \cos(2 \cdot 0) + c_2 e^{-0} \sin(2 \cdot 0) + \frac{1}{5} \] This simplifies to: \[ 0 = c_1 + \frac{1}{5} \] Solving for \(c_1\), we get \(c_1 = -\frac{1}{5}\). Now, let's differentiate \(y(t)\) with respect to \(t\) to find \(y'(t)\). \[ y^{\prime}(t) = -c_1 e^{-t} \cos(2t) - 2c_2 e^{-t} \cos(2t) + 2c_1 e^{-t} \sin(2t) - 2c_2 e^{-t}\sin(2t) \] For \(y^{\prime}(0) = 0\): \[ 0 = -c_1 \cos(0) - 2c_2 \cos(0) + 2c_1 \sin(0) - 2c_2 \sin(0) \] This simplifies to: \[ 0 = -c_1 - 2c_2 \] Substitute the previously found value of \(c_1\): \[ 0 = \frac{1}{5} - 2c_2 \] Solving for \(c_2\), we get \(c_2 = \frac{1}{10}\).

Step 6: Final solution

With the found values of \(c_1\) and \(c_2\), substitute them into the general solution: \[ y(t) = -\frac{1}{5}e^{-t}\cos(2t) + \frac{1}{10}e^{-t}\sin(2t) + \frac{1}{5} \] This is the final solution to the given initial value problem.

Key Concepts

Homogeneous Differential Equation

Understanding the concept of a homogeneous differential equation is essential in addressing many mathematical and physical problems. Simplified, these are equations set to zero and do not include any functions of the independent variable (in our case, 't'). In the exercise provided, the equation
\( y^{\textprime \textprime}(t) + 2y^{\textprime}(t) + 5y(t) = 0 \)
is already in homogeneous form, as the right-hand side is set to zero. It represents a system whose rate of change at any point depends only on its current state, not on any input or forcing function. When solved, this form of the equation gives us the complementary function, which is part of the general solution to our initial value problem.

Characteristic Equation

Characteristic equations are the backbone when it comes to solving homogeneous linear differential equations with constant coefficients. They are algebraic equations obtained by transforming the differential equation involving derivatives into a polynomial equation that doesn't — a concept crucial to our exercise. For our problem, we translate
\( y^{\textprime \textprime}(t) + 2y^{\textprime}(t) + 5y(t) = 0 \)
into the characteristic equation
\( r^2 + 2r + 5 = 0 \),
which is solved for 'r' to find the 'roots' or solutions. These roots tell us the form that the complementary solution will take. For instance, real, distinct roots suggest exponentially decaying solutions, while complex roots, as in our exercise, imply sinusoidal components due to oscillatory behavior.

Complementary Function

The complementary function is the solution to the homogeneous part of our differential equation. It's denoted as
\( y_c(t) \)
and represents the general behavior of the system without external influences. In our exercise, we derived this function after solving the characteristic equation for complex roots, resulting in
\( y_c(t) = c_1 e^{-t} \text{cos}(2t) + c_2 e^{-t} \text{sin}(2t) \).
Here, \( c_1 \) and \( c_2 \) are arbitrary constants determined by the initial conditions of the problem. The trigonometric functions reflect the oscillatory nature due to the complex roots, and the exponential term represents decay as time moves forward.

Particular Solution

On the other side of the spectrum, we have the particular solution, notated as
\( y_p(t) \),
which addresses the non-homogeneous part of the differential equation — that is the external forcing or input. For the initial value problem at hand, the function
\( H(t) \)
plays the role of this external input. Because \( H(t) \) only impacts the system within a specific range, we identify a distinct value for
\( y_p(t) \)
(within that range) to satisfy the equation when
\( y^{\textprime \textprime}(t) + 2y^{\textprime}(t) + 5y(t) = H(t) \).
In the solution provided,
\( y_p(t) = \frac{1}{5} \)
is determined for the range where \( H(t) = 1 \), illustrating how we adapt the particular solution to account for varying external conditions — a crucial understanding needed to fully grasp and solve non-homogeneous differential equations.