Problem 27 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem up to the evaluation of a convolution integral. $$y^{\prime \prime}-2 y^{\prime}+10 y=\cos 2 t, \quad y(0)=0, \quad y^{\prime}(0)=1$$

Step-by-Step Solution

Answer

The solution to the given initial-value problem is: \[y(t) = -\frac{1}{10}e^{t}\cos(4t) + \frac{1}{2} e^{t}\sin(4t) + \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t)\]

Step 1: Solve the Complementary Equation

First, we need to solve the homogeneous equation: \[ y^{\prime \prime} - 2 y^{\prime} + 10y = 0 \] The characteristic equation for this differential equation is \[ r^2 - 2r + 10 = 0 \] Solving the above quadratic equation using the quadratic formula, \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get complex conjugate roots \[ r_{1,2} = \frac{2 \pm i \sqrt{32}}{2} = 1 \pm 4i \] Thus, the complementary solution is given by \[ y_c(t) = C_1e^{t} \cos(4t) + C_2 e^{t} \sin(4t) \]

Step 2: Find the Particular Solution

Next, we will find the particular solution. Since the forcing term is \(\cos(2t)\), we will assume a particular solution of the form \[ y_p(t) = A\cos(2t) + B\sin(2t) \] Now, we need to find the first and second derivatives of \(y_p(t)\): \[ y^{\prime}_p(t) = -2A\sin(2t) + 2B\cos(2t) \] \[ y^{\prime\prime}_p(t) = -4A\cos(2t) - 4B\sin(2t) \] Substitute \(y_p(t), y^{\prime}_p(t),\) and \(y^{\prime\prime}_p(t)\) into the given non-homogeneous equation: \[ -4A\cos(2t) - 4B\sin(2t) - 2(-2A\sin(2t) + 2B\cos(2t)) + 10( A\cos(2t) + B \sin(2t) ) = \cos(2t) \] Now, combining like terms, we have: \[ (-4A + 4B + 10A)\cos(2t) + (-4B - 4A + 10B)\sin(2t) = \cos(2t) \] Equating the coefficients, we get the following system of equations: \[ 6A + 4B = 1 \] \[ -4A + 6B = 0 \] Solving the system, we find \(A = \frac{1}{10}\) and \(B = \frac{1}{5}\). Thus, the particular solution is: \[ y_p(t) = \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t) \]

Step 3: Apply the Initial Conditions

Now we need to apply the initial conditions \(y(0) = 0\) and \(y^{\prime}(0) = 1\) to find the values of \(C_1\) and \(C_2\). The general solution is the sum of the complementary and particular solutions: \[ y(t) = y_c(t) + y_p(t) = C_1e^{t}\cos(4t) + C_2 e^{t}\sin(4t) + \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t) \] Applying the first initial condition \(y(0) = 0\), we get: \[ 0 = C_1 + \frac{1}{10} \] So, \(C_1 = -\frac{1}{10}\). Now we need to apply the second initial condition, \(y^{\prime}(0) = 1\). First, let's find the derivative of our general solution: \[ y^{\prime}(t) = -\frac{1}{10}e^{t}\cos(4t) - 4 C_2 e^{t}\cos(4t) - 2 \frac{1}{10}\sin(2t) + 2 \frac{1}{5}\cos(2t) \] Applying the condition \(y^{\prime}(0) = 1\), we have: \[ 1 = -\frac{1}{10} -\frac{1}{5} => C_2 = \frac{1}{2} \] The values of \(C_1\) and \(C_2\) are now determined, and the particular solution is: \[ y(t) = -\frac{1}{10}e^{t}\cos(4t) + \frac{1}{2} e^{t}\sin(4t) + \frac{1}{10}\cos(2t) + \frac{1}{5}\sin(2t) \] Now we have found the required solution for the given initial value problem, and further evaluation using convolution integral will not be performed as per the exercise prompt.

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific conditions known as initial conditions. These are the values of the function and possibly its derivatives at a certain point, often where the independent variable equals zero. This situation is a vital aspect in dynamical systems, where the entire future behavior of the solution might be determined by these initial parameters.
Consider the equation given in the exercise:

• Differential Equation: \(y'' - 2y' + 10y = \cos 2t\)
• Initial Conditions: \(y(0) = 0\) and \(y'(0) = 1\)

These initial conditions allow us to find a specific solution to the differential equation, which would otherwise have an infinite number of potential solutions. Starting with these values helps to determine the constants that appear in the general solution. It makes the solution unique and applicable to the specific situation at hand.

Complementary Solution

The complementary solution involves solving the homogeneous version of the differential equation, where the right-hand side is zero. For our equation, the associated homogeneous differential equation is \[ y'' - 2y' + 10y = 0. \] Solving this gives us the complementary solution. It captures the part of the solution to the differential equation that doesn't get altered by the non-homogeneous part (the right-hand side, here \(\cos 2t\)). To solve, we often find the characteristic equation:

• Characteristic Equation: \(r^2 - 2r + 10 = 0\)

Using the quadratic formula, you find the roots, which might be real or complex. Here, the roots are complex conjugates: \(r = 1 \pm 4i\). This leads us to a complementary solution that involves exponentials and trigonometric functions:

• \(y_c(t) = C_1 e^t \cos(4t) + C_2 e^t \sin(4t)\)

Particular Solution

Following the complementary solution, we find the particular solution. This deals with the non-homogeneous part of the differential equation, i.e., the forcing function, which is \(\cos 2t\) in this problem. A particular solution satisfies the entire non-homogeneous differential equation, incorporating this right-hand side.
We guess a form that resembles the forcing function and calculate the coefficients by plugging back into the full equation. Here, we assume \[y_p(t) = A \cos(2t) + B \sin(2t)\] Subsequently, we take its derivatives and substitute them back into the original differential equation, equating coefficients to solve for \(A\) and \(B\). This yields:

• \(A = \frac{1}{10}\)
• \(B = \frac{1}{5}\)

The particular solution helps adjust the general solution to meet the conditions set by the non-homogeneous part of the differential equation.

Quadratic Formula

The quadratic formula is a crucial tool in finding the roots of quadratic equations like the characteristic equation we encountered when finding the complementary solution. The formula is \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], where the equation is written in the standard form \(ax^2 + bx + c = 0\).
Applying the quadratic formula to our characteristic equation \(r^2 - 2r + 10 = 0\) gives us:

• \(a = 1\), \(b = -2\), \(c = 10\)

Calculating using these values, the discriminant \(b^2 - 4ac\) results in a negative number, indicating complex roots. Specifically, these roots are \(1 \pm 4i\), leading to an oscillatory nature in the complementary solution. The quadratic formula thus provides the essential roots needed to build solutions involving exponential and trigonometric terms when the roots are complex.