Problem 86 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(67-86\). $$ \begin{array}{l}{y^{(4)}=-\cos x+8 \sin 2 x} \\ {y^{\prime \prime \prime}(0)=0, \quad y^{\prime \prime}(0)=y^{\prime}(0)=1, \quad y(0)=3}\end{array} $$

Step-by-Step Solution

Answer

The solution is \( y = 4 + x + x^2 - \cos x + \frac{1}{2} \sin 2x \).

Step 1: Find the Homogeneous Solution

For a fourth-order linear differential equation like this one, we start by finding the solution to the corresponding homogeneous equation. This is done by solving the equation where the right-hand side is zero: \( y^{(4)} = 0 \). The characteristic equation for this is \( r^4 = 0 \), which gives us roots \( r = 0, 0, 0, 0 \). Thus, the general solution to the homogeneous equation is \( y_h = C_1 + C_2x + C_3x^2 + C_4x^3 \).

Step 2: Find a Particular Solution

Now, solve for a particular solution to the non-homogeneous equation \( y^{(4)} = -\cos x + 8\sin 2x \). We assume a form for the particular solution matching the right-hand side functions. Use \( y_p = A\cos x + B\sin x + C\cos 2x + D\sin 2x \). Substitute this back into the differential equation and solve for the coefficients \( A, B, C, D \).

Step 3: Solve for Coefficients of Particular Solution

Upon substitution, the fourth derivative comes out as: \( y_p^{(4)} = A\cos x + B\sin x + 16C\cos 2x + 16D\sin 2x \). Compare it to \( -\cos x + 8\sin 2x \), giving \( A = -1 \), \( B = 0 \), \( 16C = 0 \), \( 16D = 8 \). Thus, \( D = \frac{1}{2} \). The particular solution is \( y_p = -\cos x + \frac{1}{2} \sin 2x \).

Step 4: Combine Solutions and Apply Initial Conditions

The general solution is the sum of the homogeneous and particular solutions: \( y = y_h + y_p = C_1 + C_2x + C_3x^2 + C_4x^3 - \cos x + \frac{1}{2}\sin 2x \). Use the initial conditions to solve for \( C_1, C_2, C_3, C_4 \).

Step 5: Apply Initial Conditions

Using \( y(0) = 3 \), substitute \( x = 0 \) into the general solution: \( y(0) = C_1 - 1 = 3 \). Thus, \( C_1 = 4 \). For \( y'(0) \), differentiate and substitute \( x = 0 \): \( y'(0) = C_2 = 1 \). For \( y''(0) \), differentiate again and substitute \( x = 0 \): \( y''(0) = 2C_3 - 1 = 1 \). So, \( C_3 = 1 \). For \( y'''(0) \), differentiate once more, substitute \( x = 0 \), and use the condition: \( y'''(0) = 6C_4 = 0 \), leading to \( C_4 = 0 \).

Step 6: Final Solution Assembly

This yields the specific solution: \( y = 4 + x + x^2 - \cos x + \frac{1}{2} \sin 2x \). All initial conditions are now satisfied as calculated.

Key Concepts

Homogeneous Solution

The homogeneous solution of a differential equation focuses on solving the associated equation when the right-hand side is zero. For fourth-order linear differential equations, the homogeneous form involves setting the equation to zero: \( y^{(4)} = 0 \). This simplification helps in identifying solutions that are purely dependent on the initial characteristic behavior of the differential system.
To solve it, we derive a characteristic equation from the differential equation. For our problem, the characteristic equation is \( r^4 = 0 \), indicating that all roots of \( r \) are zero with multiplicity four.

These roots help us generate the complete set of independent solutions that form the homogeneous solution. Given our roots \( r = 0, 0, 0, 0 \), the general solution in terms of constants is \( y_h = C_1 + C_2x + C_3x^2 + C_4x^3 \). This covers all basic polynomial components in our solution, accounting for the fourth-order nature.

Particular Solution

When a differential equation is non-homogeneous, it has an added function on the right-hand side, requiring us to find a particular solution that accommodates this additional element. In our example, the differential equation has a right-hand side of \( -\cos x + 8 \sin 2x \).
To find a particular solution, we assume a trial solution structure similar to the non-homogeneous part. We propose a solution form like \( y_p = A\cos x + B\sin x + C\cos 2x + D\sin 2x \). This mirrors the sine and cosine terms in the original equation.
Substituting \( y_p \) into the original differential equation leads us to determine the coefficients \( A, B, C, \) and \( D \) for fitting the given terms. After simplifying, we find \( A = -1 \), \( B = 0 \), \( 16C = 0 \), and \( D = \frac{1}{2} \), yielding \( y_p = -\cos x + \frac{1}{2} \sin 2x \). This specific combination fits the non-homogeneous part, creating a balanced equation.

Initial Value Problem

An initial value problem involves finding a specific solution to a differential equation that satisfies given initial conditions. These conditions help hone in on one exact solution from the general set defined by \( y = y_h + y_p \).
In this exercise, we have initial conditions defined at \( x = 0 \):

• \( y(0) = 3 \)
• \( y'(0) = 1 \)
• \( y''(0) = 1 \)
• \( y'''(0) = 0 \)

These conditions are used progressively to determine the constants \( C_1, C_2, C_3, \) and \( C_4 \) in the general solution.
By substituting these initial condition values into the general solution and its derivatives, we progressively solve for each constant.

• \( C_1 \) is determined by satisfying \( y(0) = 3 \) leading to \( C_1 = 4 \).
• \( C_2 \) is found when aligning with \( y'(0) = 1 \).
• \( C_3 \) uses \( y''(0) = 1 \) to solve for it.
• \( C_4 \) involves \( y'''(0) = 0 \), finding it as 0.

Together, these constants shape the full solution, aligning the function to meet all initial conditions.

Characteristic Equation

Characterizing the nature of solutions in a differential equation begins with the characteristic equation. This is essentially an algebraic equation derived to help determine the behavior of a differential equation's solutions, particularly in its homogeneous form.
For our fourth-order differential equation \( y^{(4)} = 0 \), the characteristic equation is formulated as \( r^4 = 0 \).
Characteristic equations typically reveal crucial information such as the types of roots (real or complex) and their multiplicities, helping in constructing a general form for the solution.

• In this case, our roots are \( r = 0, 0, 0, 0 \).
• Each root's multiplicity drives the polynomials \( 1, x, x^2, x^3 \) in the solution.....

Such roots establish the framework for combining polynomials into the general homogeneous solution \( y_h = C_1 + C_2x + C_3x^2 + C_4x^3 \).
With this, the characteristic equation plays a pivotal role in penciling the basic building blocks of the solution space, giving direction for further calculations.