Problem 61 - Essential Calculus

Question

Find the particular solution of the differential equation that satisfies the initial conditions. $$ f^{\prime \prime}(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right), \quad f(0)=1, f^{\prime}(0)=0 $$

Step-by-Step Solution

Answer

The particular solution of given differential equation is \(f(x) = \frac{1}{2}e^{x} + \frac{1}{2}e^{-x} + 1\).

Step 1: Identify the type of differential equation

The given differential equation is of the form \(f''(x) = g(x)\), where \(g(x) = \frac{1}{2}(e^x + e^{-x})\). This is a second order non-homogeneous differential equation.

Step 2: Determine the complementary solution

The complementary solution, \(f_c(x)\), of the equation is the solution of the homogeneous form, i.e when \(g(x) = 0\). So, we solve the equation \(f''(x) = 0\). The general solution of this differential equation is \(f_c(x) = Ax + B\), where A and B are constants.

Step 3: Find the particular solution

To solve for particular solution, we make a guess based on the right hand side of the equation. In this case, it is of the form \(Ce^{x} + De^{-x}\). Differentiating this guess once gives \(Ce^{x} - De^{-x}\) and differentiating again gives \(Ce^{x} + De^{-x}\). Equating this to our right hand side \(\frac{1}{2}(e^x + e^{-x})\), we get \(C = D = 1/2.\) Therefore, the particular solution, \(f_p(x)\), is \(\frac{1}{2}e^{x} + \frac{1}{2}e^{-x}\).

Step 4: Form the general solution

The general solution is the sum of the complementary and the particular solution, giving us \(f(x) = Ax + B + \frac{1}{2}e^{x} + \frac{1}{2}e^{-x}\).

Step 5: Apply the initial conditions to find A and B

Apply the initial conditions, \(f(0) = 1\) and \(f'(0) = 0\). After plugging these values into \(f(x)\) and \(f'(x)\) respectively, we obtain \(A = 0\) and \(B = 1\).

Key Concepts

Particular Solution

When tackling differential equations, finding a particular solution is crucial. A particular solution addresses the non-homogeneity in the equation. Simply put, it's a solution that satisfies the differential equation with the specific function on the right side of the equals sign.

In the step-by-step solution above, the goal was to find a particular solution to this non-homogeneous equation:
\[ f''(x) = \frac{1}{2}(e^x + e^{-x}) \]The key idea is to "guess" a solution based on the format of the non-homogeneous term, \( g(x) \). Here, since the right-hand side features exponentials \( e^x \) and \( e^{-x} \), a reasonable guess was:

• \( f_p(x) = Ce^{x} + De^{-x} \)

Through differentiation and equation matching, specific values for \(C\) and \(D\) were found, allowing determination of:

• \( f_p(x) = \frac{1}{2}e^{x} + \frac{1}{2}e^{-x} \)

Finding the particular solution isn't just guessing; it involves thoughtful selection based on the structure of the equation, followed by verification.

Initial Conditions

Initial conditions are essential for determining a specific solution from the infinite possibilities provided by a differential equation's general solution. They are essentially values known at the "start" point, often used to solve constants in the solution.

In our example, the initial conditions given were:

• \( f(0) = 1 \)
• \( f'(0) = 0 \)

These provide critical boundary values necessary to solve for constants \(A\) and \(B\) in the general solution, making the broad, general solution particular to a single curve that fits these conditions.

By applying \(f(0) = 1\) and \(f'(0) = 0\) to:

• \( f(x) = Ax + B + \frac{1}{2}e^{x} + \frac{1}{2}e^{-x} \)
• \( f'(x) = A + \frac{1}{2}e^{x} - \frac{1}{2}e^{-x} \)

we determine \(A = 0\) and \(B = 1\). Initial conditions let us pinpoint this specific solution that fits the system's physical context or initial setup.

Non-Homogeneous Differential Equation

A non-homogeneous differential equation includes a term independent from the function and its derivatives, introducing an external component driving change in the system.

For example, our equation:

• \( f''(x) = \frac{1}{2}(e^x + e^{-x}) \)

is non-homogeneous because of \(\frac{1}{2}(e^x + e^{-x})\), differentiating it from its homogeneous counterpart, \( f''(x) = 0 \).

Steps to solve such equations include:

• Finding the complementary solution, \( f_c(x) \), for the homogeneous part.
• Determining a particular solution, \( f_p(x) \), for the non-homogeneous part.
• Combining these to form a general solution: \( f(x) = f_c(x) + f_p(x) \).

Non-homogeneous differential equations often model real-world scenarios where external forces or inputs influence the system, thus making them more realistic in practical applications.