Problem 35 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. \(y^{\prime \prime \prime}-2 y^{\prime \prime}+y^{\prime}=2-24 e^{x}+40 e^{5 x}, y(0)=\frac{1}{2}, y^{\prime}(0)=\frac{5}{2}\), \(y^{\prime \prime}(0)=-\frac{9}{2}\)

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{1}{2} + 801 e^{x} - \frac{1601}{2} xe^{x} + 40e^{5x} \).

Step 1: Write the Form of the General Solution

The given equation is a non-homogeneous linear differential equation with constant coefficients:\[ y''' - 2y'' + y' = 2 - 24e^{x} + 40e^{5x} \]First, solve the homogeneous equation:\[ y''' - 2y'' + y' = 0 \] Find the complementary solution (\(y_c\)) by solving the characteristic equation.

Step 2: Solve the Characteristic Equation

The characteristic equation for the homogeneous equation is:\[ r^3 - 2r^2 + r = 0 \]Factoring out an \(r\), gives:\[ r(r^2 - 2r + 1) = 0 \]This simplifies to \(r=0\) and \( (r-1)^2 = 0 \), giving roots \(r = 0, 1, 1\). The solution to the homogeneous equation is:\[ y_c = C_1 + C_2 e^{x} + C_3 xe^{x} \]

Step 3: Determine the Particular Solution

Next, assume a particular solution, \(y_p\), due to the non-homogeneous part.We try solutions of the form:\[ y_p = A + Be^{x} + Ce^{5x} \]Derive \(y_p', y_p''\), and \(y_p'''\) and substitute into:\[ y''' - 2y'' + y' = 2 - 24e^{x} + 40e^{5x} \] and solve for the coefficients \(A, B, C\).

Step 4: Solve for Coefficients in Particular Solution

The derivatives of \(y_p\) are:- \(y_p' = Be^{x} + 5Ce^{5x}\)- \(y_p'' = Be^{x} + 25Ce^{5x}\)- \(y_p''' = Be^{x} + 125Ce^{5x}\)Substitute these into the original equation:\[ Be^{x} + 125Ce^{5x} - 2(Be^{x} + 25Ce^{5x}) + (Be^{x} + 5Ce^{5x}) = 2 - 24e^{x} + 40e^{5x} \]Simplifying gives:\[ -24e^{x} + 40e^{5x} = 2 - 24e^{x} + 40e^{5x} \]Thus, \(A = 2, B = 0, C = 40\) so:\[ y_p = 2 + 40e^{5x} \]

Step 5: Combine Solutions

The general solution is:\[ y = y_c + y_p = C_1 + C_2 e^{x} + C_3 xe^{x} + 2 + 40e^{5x} \]

Step 6: Apply Initial Conditions

Apply the initial conditions to solve for the constants:1. \(y(0)=\frac{1}{2}\): \[ C_1 + 2 = \frac{1}{2} \rightarrow C_1 = -\frac{3}{2} \]2. \(y'(0)=\frac{5}{2}\): \[ y'(x) = C_2 e^{x} + C_3 e^{x} + C_3 xe^{x} + 200e^{5x} \] \[ \text{At } x=0, y'(0) = C_2 + C_3 + 200 = \frac{5}{2} \]3. \(y''(0)=-\frac{9}{2}\): \[ y''(x) = C_2 e^{x} + 2C_3 e^{x} + C_3 xe^{x} + 1000e^{5x} \] \[ \text{At } x=0, y''(0) = C_2 + 2C_3 + 1000 = -\frac{9}{2} \]Solve these equations to find \(C_2\) and \(C_3\).

Step 7: Solve for Constants

From the equation system:- \(C_1 = -\frac{3}{2}\)- \(C_2 + C_3 + 200 = \frac{5}{2} \) implies \(C_2 + C_3 = \frac{5}{2} - 200 = -\frac{395}{2}\)- \(C_2 + 2C_3 + 1000 = -\frac{9}{2} \) implies \(C_2 + 2C_3 = -\frac{9}{2} - 1000 = -\frac{2009}{2}\)Solving these gives \(C_2 = 801\) and \(C_3 = -\frac{1601}{2}\).

Step 8: Write the Final Solution

Substitute the constants back into the general solution:\[ y(x) = -\frac{3}{2} + 801 e^{x} - \frac{1601}{2} xe^{x} + 2 + 40e^{5x} \]Thus, the final solution to the initial-value problem is:\[ y(x) = \frac{1}{2} + 801 e^{x} - \frac{1601}{2} xe^{x} + 40e^{5x} \]

Key Concepts

Initial-Value Problems

An initial-value problem involves a differential equation along with specified values at the initial point of the variables involved. These particular values, known as initial conditions, allow us to find a unique solution to a differential equation.
The given problem involves the third-order differential equation:
\( y''' - 2y'' + y' = 2 - 24e^{x} + 40e^{5x} \), with initial conditions \( y(0)=\frac{1}{2} \), \( y'(0)=\frac{5}{2} \), and \( y''(0)=-\frac{9}{2} \).
The goal is to find the function \( y(x) \) that satisfies both this differential equation and the initial conditions.

• Initial conditions provide specific values of the function and its derivatives at a starting point.
• They are essential to distinguish between multiple possible solutions.

Understanding and applying initial-value problems is vital in predicting and modeling physical phenomena, where specific conditions are always present at the start.

Particular Solution

The particular solution of a differential equation deals with finding a specific solution to the non-homogeneous equation, differentiating it from the general solution.
In this equation, \( y''' - 2y'' + y' = 2 - 24e^{x} + 40e^{5x} \), we aim to find \( y_p \):

• The particular solution is influenced by the non-homogeneous part (\(2 - 24e^{x} + 40e^{5x}\)).
• We assume a form for \( y_p \), usually involving constants and functions matching those in the problem.

For this exercise:
Suppose \( y_p = A + Be^{x} + Ce^{5x} \).
By substituting and matching coefficients:
- The derivatives are substituted back into the equation, leading us to solve for the coefficients \(A, B, C\).
Thus, \( y_p = 2 + 40e^{5x} \) based on the particularities here.

Complementary Solution

The complementary solution is the part of the general solution to a homogeneous differential equation.
For the homogeneous counterpart of our differential equation:
\( y''' - 2y'' + y' = 0 \)
We found the complementary solution, \( y_c \), by solving the characteristic equation.

• The roots of the characteristic equation provide us the basis for \( y_c \).

The characteristic equation:
\( r^3 - 2r^2 + r = 0 \), factored to get:\( r(r-1)^2 = 0 \), yields roots \( r = 0, 1, 1 \).
This gives:
\( y_c = C_1 + C_2 e^{x} + C_3 xe^{x} \).
The complementary solution represents all the possible solutions to the homogeneous equation and forms a foundation for the overall solution.

Characteristic Equation

A characteristic equation is essential for solving linear homogeneous differential equations with constant coefficients.
For the differential equation \( y''' - 2y'' + y' = 0 \), the associated characteristic equation is derived by replacing each derivative with powers of \( r \):

• The characteristic equation reflects the algebraic form: \( r^3 - 2r^2 + r = 0 \).
• Solving it gives us the vital roots \( r \) which inform on the nature of the complementary solution.

Through factoring \( r \) out, it becomes:
\( r(r^2 - 2r + 1) = 0 \), which further simplifies to \( r(r-1)^2 = 0 \) with roots \( r = 0, 1, 1 \).
Each root corresponds to a component in the complementary solution:
- Different roots give distinct solutions.
These steps are crucial as they determine how the general solution is structured and how specific solutions to these differential equations can be articulated.