Question

Find the \(\gamma_{p}\) and the \(y_{c \prime}\) the general solution, and the definite solution of each of the following: $$y^{\prime \prime}(t)+9 y=3 ; y(0)=1, y^{\prime}(0)=3$$

Short answer

The definite solution is \(y(t) = \frac{2}{3} \cos(3t) + \sin(3t) + \frac{1}{3}\).

Step-by-step solution

Identify the Type of Equation

The given differential equation is \(y''(t) + 9y = 3\). This is a second-order linear differential equation with constant coefficients.

Find the Homogeneous Solution

To find the general solution, start with the corresponding homogeneous equation: \(y''(t) + 9y = 0\). The characteristic equation for this is \(r^2 + 9 = 0\). Solving for \(r\), we find \(r = \pm 3i\). Therefore, the homogeneous solution is \(y_h(t) = c_1 \cos(3t) + c_2 \sin(3t)\).

Find a Particular Solution

Since the non-homogeneous term is a constant (3), choose a particular solution of the form \(y_p(t) = A\), where \(A\) is a constant. Applying this to the differential equation, \(0 + 9A = 3\), solving gives \(A = \frac{1}{3}\). Hence, the particular solution is \(y_p(t) = \frac{1}{3}\).

Form the General Solution

The general solution is the sum of the homogeneous and particular solutions: \(y(t) = y_h(t) + y_p(t) = c_1 \cos(3t) + c_2 \sin(3t) + \frac{1}{3}\).

Apply Initial Conditions

To find the definite solution, use the initial conditions \(y(0) = 1\) and \(y'(0) = 3\). Substituting \(t = 0\) in the general solution, \(c_1 + \frac{1}{3} = 1\) leads to \(c_1 = \frac{2}{3}\). Differentiating the general solution gives \(y'(t) = -3c_1 \sin(3t) + 3c_2 \cos(3t)\). At \(t = 0\), \(3c_2 = 3\) implying \(c_2 = 1\).

Write the Definite Solution

Substitute \(c_1 = \frac{2}{3}\) and \(c_2 = 1\) into the general solution to obtain the definite solution: \(y(t) = \frac{2}{3} \cos(3t) + \sin(3t) + \frac{1}{3}\).

Key concepts

Second-Order Differential Equation

A second-order differential equation is an equation that involves the unknown function and its derivatives up to the second order. In our task, we are dealing with the equation \(y''(t) + 9y = 3\). Here, \(y''(t)\) stands for the second derivative of \(y\) with respect to \(t\). Second-order differential equations are fundamental in describing various physical phenomena, such as oscillations and waves.
These equations can be linear or non-linear. In this case, the equation is linear because the unknown function \(y\) and its derivatives are linear. The presence of constant coefficients, like \(9\) in our equation, indicates that the equation has constant coefficients, which simplifies the process of finding solutions. Identifying the order and type of the differential equation is crucial as it guides us to the appropriate methods for finding solutions.

Homogeneous Solution

To solve a second-order differential equation, we first seek the homogeneous solution, \(y_h(t)\). This involves setting the non-homogeneous part (\(3\), in our example) to zero, resulting in \(y''(t) + 9y = 0\). This step focuses only on the natural behavior of the system without external influences.
The method of characteristic equations is used here by translating the differential equation into its characteristic form. For \(y''(t) + 9y = 0\), the characteristic equation becomes \(r^2 + 9 = 0\). Solving this gives complex roots \(r = \pm 3i\), indicating oscillatory behavior.
The general homogeneous solution is then formed using these roots, resulting in \(y_h(t) = c_1 \cos(3t) + c_2 \sin(3t)\). Here, \(c_1\) and \(c_2\) are arbitrary constants determined by initial or boundary conditions.

Particular Solution

Next, we look for the particular solution, \(y_p(t)\), which complements the homogeneous solution to address the complete equation including the non-homogeneous term (\(3\) in our equation \(y''(t) + 9y = 3\)). Since the non-homogeneous term is a constant, we assume a particular solution of this form: \(y_p(t) = A\).
Substituting \(y_p = A\) into the differential equation, we attain \(0 + 9A = 3\). Solving for \(A\) gives \(A = \frac{1}{3}\). Consequently, the particular solution \(y_p(t)\) is \(\frac{1}{3}\).
This step ensures that the specific behavior imposed by the non-homogeneous part of the equation is accounted for, providing a solution that covers all aspects of the equation.

Initial Conditions

Finally, we use initial conditions to find the definite solution and determine the arbitrary constants in the general solution. In our case, two initial conditions are provided: \(y(0) = 1\) and \(y'(0) = 3\).

• By substituting \(t = 0\) into the general solution \(y(t) = c_1 \cos(3t) + c_2 \sin(3t) + \frac{1}{3}\), the equation \(c_1 + \frac{1}{3} = 1\) is obtained, leading to \(c_1 = \frac{2}{3}\).
• Next, differentiating the general solution gives \(y'(t) = -3c_1 \sin(3t) + 3c_2 \cos(3t)\). Substituting \(t = 0\) and applying the condition \(y'(0) = 3\), we find \(3c_2 = 3\) or \(c_2 = 1\).

Substituting \(c_1 = \frac{2}{3}\) and \(c_2 = 1\) into the general solution gives the exact solution: \(y(t) = \frac{2}{3} \cos(3t) + \sin(3t) + \frac{1}{3}\). Initial conditions are crucial because they provide the necessary specifics to refine the general solution to one precise solution that fits the scenario described.