Problem 17 - Calculus: Early Transcendentals

Question

Solve the initial value problem. $$y^{\prime \prime}-y=3 t+5, y(0)=0, y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

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

Step 1: Solve the Homogeneous Equation

First, consider the homogeneous part of the differential equation:\[ y'' - y = 0 \]Assume a solution of the form \( y_h = e^{rt} \). Substitute into the equation:\[ r^2 e^{rt} - e^{rt} = 0 \]Cancel \( e^{rt} \) since it is not zero:\[ r^2 - 1 = 0 \]The roots are \( r = 1 \) and \( r = -1 \). Thus, the general solution to the homogeneous equation is:\[ y_h = C_1 e^{t} + C_2 e^{-t} \].

Step 2: Solve the Particular Equation

Next, solve the non-homogeneous equation:\[ y'' - y = 3t + 5 \]Assume a particular solution of the form:\[ y_p = At + B \]Take the derivatives:\[ y_p' = A \]\[ y_p'' = 0 \]Substitute into the differential equation:\[ 0 - (At + B) = 3t + 5 \]This simplifies to:\[ -At - B = 3t + 5 \]Equate coefficients to solve for \( A \) and \( B \):- \(-A = 3 \rightarrow A = -3\)- \(-B = 5 \rightarrow B = -5\)So, the particular solution is:\[ y_p = -3t - 5 \].

Step 3: Form the General Solution

Combine the solutions from Step 1 and Step 2 to form the general solution:\[ y = y_h + y_p = C_1 e^t + C_2 e^{-t} - 3t - 5 \].

Step 4: Apply Initial Conditions

Use the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \).1. Apply \( y(0) = 0 \): \[ C_1 e^0 + C_2 e^0 - 3(0) - 5 = 0 \] \[ C_1 + C_2 - 5 = 0 \] \[ C_1 + C_2 = 5 \] \( \rightarrow (1) \)2. Find \( y'(t) \): \[ y' = C_1 e^t - C_2 e^{-t} - 3 \] \( y'(0) = 0 \): \[ C_1 - C_2 - 3 = 0 \] \[ C_1 - C_2 = 3 \] \( \rightarrow (2) \)Solve equations (1) and (2): Adding: \[ 2C_1 = 8 \rightarrow C_1 = 4 \] Substituting \( C_1 \) in (1): \[ 4 + C_2 = 5 \rightarrow C_2 = 1 \].

Step 5: Write the Final Solution

Substitute the values of \( C_1 \) and \( C_2 \) into the general solution:\[ y = 4 e^t + 1 e^{-t} - 3t - 5 \]Thus, the solution satisfying the initial conditions is:\[ y(t) = 4 e^t + e^{-t} - 3t - 5 \].

Key Concepts

Homogeneous Equation

A homogeneous equation is an important concept in differential equations. It refers to an equation where the non-homogeneous part, usually a function on the right side, is zero. For example, in the given problem, the homogeneous equation is: \[ y'' - y = 0 \]In this context, our job is to find solutions for this equation assuming there's no external force or input, signified by the right side being zero.To solve it, we assume a solution of the form \( y_h = e^{rt} \). Substituting this form into the equation allows us to cancel out common factors and find a characteristic equation that looks like:\[ r^2 - 1 = 0 \]Solving for \( r \) gives us the roots, which are \( r = 1 \) and \( r = -1 \). This results in a general solution:\[ y_h = C_1 e^t + C_2 e^{-t} \]Here, \( C_1 \) and \( C_2 \) are constants that we will later determine using initial conditions. This solution to the homogeneous equation represents the natural behavior of the system without external inputs.

Particular Solution

To solve a non-homogeneous differential equation, we need to find a particular solution that satisfies the equation with its specific non-zero terms. In our problem, the non-homogeneous part of the equation is:\[ y'' - y = 3t + 5 \] The term \( 3t + 5 \) signifies an external influence or input into the system. To find a particular solution \( y_p \), it is wise to guess a polynomial form, given that the input is a polynomial. Hence, for this equation, we assume:\[ y_p = At + B \]Deriving, we have:

• \( y_p' = A \)
• \( y_p'' = 0 \)

Substituting these into the original differential equation leads us to a system where we solve:

• \(-At - B = 3t + 5\)
• This results in \( A = -3 \) and \( B = -5 \)

As such, the particular solution becomes:\[ y_p = -3t - 5 \]Adding this particular solution to the homogeneous solution will help us form the general solution.

Initial Value Problem

Initial value problems involve finding a specific solution for a differential equation given certain initial conditions. For our problem, the initial conditions are:

• \( y(0) = 0 \)
• \( y'(0) = 0 \)

These conditions provide specific values at a given point, often critical in applications where the starting state of a system is defined. When applying these conditions, it helps to substitute them into the general solution and its derivative to find the values of any constants present.In this exercise, the conditions are applied to the combined general solution from the homogeneous and particular solution to form a complete solution that fits the problem's constraints.Given initial conditions:- By substituting into the general solution:\[ C_1 + C_2 - 5 = 0 \] (Eq. 1) and\[ C_1 - C_2 = 3 \] (Eq. 2)We solve these simultaneous equations to find the constants \( C_1 \) and \( C_2 \), and these constants tailor the general solution to fit initial conditions.

General Solution

The general solution of a differential equation is a combination of both the homogeneous and particular solutions. It represents a family of solutions applicable to varied conditions. In our case, the general solution takes the form:\[ y = y_h + y_p = C_1 e^t + C_2 e^{-t} - 3t - 5 \]This solution combines the natural response of the system, \( y_h \), and the particular solution \( y_p \) which arises from the specific external inputs described in the equation. The constants \( C_1 \) and \( C_2 \), which arise from solving initial conditions, tailor the general solution for our specific problem:

• We found \( C_1 = 4 \) and \( C_2 = 1 \) from the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \).
• By substituting these into our general solution:

\[ y(t) = 4 e^t + e^{-t} - 3t - 5 \]This final form gives us the specific behavior of the system at any time \( t \), satisfying the initial value constraints.