Question

Find the \(\gamma_{p}\) and the \(y_{c \prime}\) the general solution, and the definite solution of each of the following: $$y^{\prime \prime}(t)+3 y^{\prime}(t)-4 y=12 ; y(0)=2, y^{\prime}(0)=2$$

Short answer

The definite solution is \( y(t) = \frac{22}{5}e^{t} + \frac{3}{5}e^{-4t} - 3 \).

Step-by-step solution

Find the Complementary Function (CF)

To solve the differential equation, first, find the complementary function. Start with the characteristic equation obtained from the differential equation without the constant term:\[ m^2 + 3m - 4 = 0 \]Solve this quadratic equation to find the roots. Use the quadratic formula:\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \(a = 1\), \(b = 3\), and \(c = -4\):\[ m = \frac{-3 \pm \sqrt{3^2 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2} \]\[ m = \frac{-3 \pm 5}{2} \]This gives roots \( m_1 = 1, m_2 = -4 \). The complementary function is:\[ y_c(t) = C_1e^{t} + C_2e^{-4t} \]

Find the Particular Solution (PS)

Next, find the particular solution, \(y_p(t)\). For the non-homogeneous differential equation\[y''(t) + 3y'(t) - 4y = 12,\]we choose a particular solution of the form \(y_p(t) = A\), because the right-hand side is a constant. Substitute into the equation:\[y'' = 0, \; y' = 0, \; y = A\]\[-4A = 12 \]Solve for \(A\):\[A = -3\].Thus, the particular solution is:\[ y_p(t) = -3 \].

Write the General Solution

Combine the complementary function and the particular solution to find the general solution of the differential equation:\[ y(t) = y_c(t) + y_p(t) = C_1e^{t} + C_2e^{-4t} - 3 \]

Apply Initial Conditions

Use the initial conditions to solve for \(C_1\) and \(C_2\). We know:\[ y(0) = 2, \; y'(0) = 2 \].Substitute \(t = 0\) into the general solution:\[ 2 = C_1e^{0} + C_2e^{0} - 3 \]\[ 2 = C_1 + C_2 - 3 \]\[ C_1 + C_2 = 5 \; (\text{Equation 1})\]Now, find \(y'(t)\) using the general solution:\[ y'(t) = C_1e^{t} - 4C_2e^{-4t} \]At \(t = 0\):\[ 2 = C_1 - 4C_2 \; (\text{Equation 2})\]

Solve System of Equations

Solve the system of equations from the initial conditions:1. \( C_1 + C_2 = 5 \)2. \( C_1 - 4C_2 = 2 \)Add these equations:\[2C_1 - 3C_2 = 7 \]From (1): \( C_1 = 5 - C_2 \).Substitute into the second equation:\[ (5 - C_2) - 4C_2 = 2 \]\[ 5 - 5C_2 = 2 \]\[ 5C_2 = 3 \]\[ C_2 = \frac{3}{5} \]Substitute \(C_2 = \frac{3}{5}\) into Equation 1:\[ C_1 = 5 - \frac{3}{5} = \frac{22}{5} \]

Write the Definite Solution

With \(C_1\) and \(C_2\) known, substitute back into the general solution:\[ y(t) = \frac{22}{5}e^{t} + \frac{3}{5}e^{-4t} - 3 \]This is the definite solution that satisfies the initial conditions.

Key concepts

Complementary Function

When solving a linear differential equation, the goal is to find the general solution. A key step involves determining the complementary function (CF), which represents the homogeneous solution (i.e., the solution of the equation without the non-homogeneous term).
To derive the CF, we start with the characteristic equation, a vital part derived from the original differential equation. Here, the differential equation is:\[ y^{\prime\prime}(t) + 3y^{\prime}(t) - 4y = 12. \]Without the constant term (12), the characteristic equation is:\[ m^2 + 3m - 4 = 0. \]Solving this quadratic equation, typically using the quadratic formula, provides the roots \( m_1 \) and \( m_2 \).
In this instance, the roots are \( m_1 = 1 \) and \( m_2 = -4 \). The general form of the complementary function, therefore, is a combination of these exponential terms:

• \( y_c(t) = C_1 e^{m_1 t} + C_2 e^{m_2 t} \)
• \( y_c(t) = C_1 e^{t} + C_2 e^{-4t} \)

Understanding the complementary function is essential, as it forms the foundational part of the general solution, alongside the particular solution.

Particular Solution

For a non-homogeneous differential equation, the particular solution (PS) complements the complementary function to form the general solution.
The particular solution addresses the non-homogeneous component (in our case, the constant 12).
Finding the PS requires an educated guess or an approach known as the method of undetermined coefficients. This involves choosing a trial solution that matches the form of the non-homogeneous component.
Given:\[ y^{\prime\prime}(t) + 3y^{\prime}(t) - 4y = 12, \]we guess \( y_p(t) = A \), assuming a constant solution since the non-homogeneous term is a constant (12).
Substituting into the differential equation, we simplify and solve for \( A \):

• \( -4A = 12 \)
• \( A = -3 \)

This gives us the particular solution:\[ y_p(t) = -3. \]The precise choice of \( y_p(t) \) depends on the non-homogeneous term's nature. By adding \( y_p(t) = -3 \) to the complementary function, we start forming the general solution.

General Solution

The general solution of a differential equation combines both the complementary function (CF) and the particular solution (PS) to solve the entire equation.
The CF addresses the homogeneous part, while the PS accounts for the non-homogeneous part. Here's the setup:

• Complementary Function: \( y_c(t) = C_1 e^t + C_2 e^{-4t} \)
• Particular Solution: \( y_p(t) = -3 \)

Thus, the general solution will be:\[ y(t) = y_c(t) + y_p(t) = C_1 e^t + C_2 e^{-4t} - 3. \]The constants \( C_1 \) and \( C_2 \) are not yet determined until we apply the initial conditions. In practice, having the general solution allows for adaptation to specific conditions or constraints, such as initial values, that adapt the equation to real-world scenarios.

Initial Conditions

Initial conditions are specific values that allow the determination of the unique constants in the general solution to tailor it to particular scenarios.
For our differential equation and its general solution:\[ y(t) = C_1 e^t + C_2 e^{-4t} - 3, \]we have initial conditions given as \( y(0) = 2 \) and \( y^{\prime}(0) = 2 \). These conditions specify the state of the system at a specific time \( t = 0 \).
By substituting these values into our general solution, we set up equations to solve for \( C_1 \) and \( C_2 \).

• Let \( t = 0 \):
  \( 2 = C_1 + C_2 - 3 \), resulting in \( C_1 + C_2 = 5. \)
• Calculate the derivative \( y^{\prime}(t) = C_1 e^t - 4C_2 e^{-4t} \) and substitute \( t = 0 \):
  \( 2 = C_1 - 4C_2 \).

Now, we solve these equations simultaneously to find \( C_1 \) and \( C_2 \):

• From equation \( C_1 + C_2 = 5 \)
• Equation \( C_1 - 4C_2 = 2 \)

The solution gives \( C_1 = \frac{22}{5} \) and \( C_2 = \frac{3}{5} \).
With these values, substitute back into the general solution to find the definite solution that satisfies the initial conditions.