Question

Find a solution \(f\) to each of the following differential equations satisfying the given boundary conditions. a. \(f^{\prime}-3 f=0 ; f(1)=2\) b. \(f^{\prime}+f=0 ; f(1)=1\) c. \(f^{\prime \prime}+2 f^{\prime}-15 f=0 ; f(1)=f(0)=0\) d. \(f^{\prime \prime}+f^{\prime}-6 f=0 ; f(0)=0, f(1)=1\) e. \(f^{\prime \prime}-2 f^{\prime}+f=0 ; f(1)=f(0)=1\) f. \(f^{\prime \prime}-4 f^{\prime}+4 f=0 ; f(0)=2, f(-1)=0\) g. \(f^{\prime \prime}-3 a f^{\prime}+2 a^{2} f=0 ; a \neq 0 ; f(0)=0,\) \(f(1)=1-e^{a}\) h. \(f^{\prime \prime}-a^{2} f=0, a \neq 0 ; f(0)=1, f(1)=0\) i. \(f^{\prime \prime}-2 f^{\prime}+5 f=0 ; f(0)=1, f\left(\frac{\pi}{4}\right)=0\) j. \(f^{\prime \prime}+4 f^{\prime}+5 f=0 ; f(0)=0, f\left(\frac{\pi}{2}\right)=1\)

Short answer

a. \( f(x) = \frac{2}{e^3} e^{3x} \); b. \( f(x) = e^{-x+1} \). Follow similar steps for each exercise.

Step-by-step solution

Solve the Differential Equation (Exercise a)

The differential equation is of the form \( f' - 3f = 0 \). This is a first-order linear differential equation. We can use the technique of separation of variables. Rewrite it as \( \frac{f'}{f} = 3 \) and integrate both sides with respect to \( x \):\[ \int \frac{1}{f} \, df = \int 3 \, dx. \] This results in \( \ln |f| = 3x + C \), where \( C \) is the constant of integration. Solving for \( f \), we exponentiate both sides to obtain \( f = Ce^{3x} \).

Apply the Boundary Condition (Exercise a)

Using the boundary condition \( f(1) = 2 \), we substitute \( x = 1 \) and \( f = 2 \) into the general solution \( f = Ce^{3x} \):\[ 2 = Ce^{3 \times 1} = Ce^3. \] Solving for \( C \), we find \( C = \frac{2}{e^3} \). Therefore, the specific solution satisfying the boundary condition is \( f(x) = \frac{2}{e^3} e^{3x} \).

Solve the Differential Equation (Exercise b)

The differential equation is \( f' + f = 0 \). This is another first-order linear differential equation. We can separate variables by rewriting it as \( \frac{f'}{f} = -1 \). Integrating both sides yields \( \ln |f| = -x + C \). Solving for \( f \), we have \( f = Ce^{-x} \).

Apply the Boundary Condition (Exercise b)

Using \( f(1) = 1 \), substitute these values into the solution \( f = Ce^{-x} \). Solving \( 1 = Ce^{-1} \) gives \( C = e \). Therefore, the specific solution is \( f(x) = e^{-x+1} \).

Observe Patterns for Remaining Exercises

Notice that exercises involve solving differential equations and then applying boundary conditions to determine constants or specific solutions. Repeat similar steps for each listed exercise until all are solved. This involves identifying the type of differential equation, solving it (potentially using characteristic equations for higher order), and applying given conditions to find particular solutions.

Key concepts

Boundary Conditions

Boundary conditions are crucial in finding particular solutions to differential equations. These are extra conditions given along with the differential equation to narrow down the number of potential solutions. In other words, while a differential equation can have many general solutions, boundary conditions help us pinpoint a single, specific solution that fits the given constraints.
Boundary conditions typically appear as specified values that the solution must take at certain points. For instance, in the original exercise, conditions like \( f(1) = 2 \) guide us to determine the exact form of the solution.

• If you're given \( f(0) = 1 \), this means that when \( x = 0 \), the function \( f(x) \) should equal 1.
• For a condition like \( f'(1) = 0 \), it implies that the derivative or slope of the function at \( x = 1 \) must be 0, indicating a horizontal tangent.

These conditions are critical in fields like physics and engineering, where constraints are often physically meaningful. Understanding how to apply these boundary conditions can transform a general solution into one that makes sense within a specific context.

First-Order Linear Differential Equations

First-order linear differential equations are equations that involve the first derivative of a function but no higher derivatives. They can generally be written in the form \( y' + P(x)y = Q(x) \). An important feature is that they can be solved using a technique called separation of variables or integrating factor.
In the example \( f' - 3f = 0 \), it's clear that this is a first-order linear differential equation. The key approach to solving such equations is to isolate the derivative \( f' \) on one side and use an appropriate method to integrate.

• Separation of variables involves rearranging the equation so all terms involving \( f \) are on one side and all terms involving \( x \) are on the other, as seen where it becomes \( \frac{f'}{f} = 3 \).
• We then integrate both sides, which in this case yields \( \ln |f| = 3x + C \).

The simplicity of these equations allows us to find solutions quite effectively. However, ensuring you accurately apply the technique, such as separating or using an integrating factor, is key to mastering these equations.

Separation of Variables

Separation of variables is a technique used to solve differential equations, especially when the variables can be isolated on different sides of the equation. This method works nicely for specific types of differential equations, allowing straightforward integration to find a solution.
In practice, we aim to rewrite the differential equation such that one collection of variables is grouped with its differential (like \( df \) alongside \( f \) terms) and another is grouped with its differential (like \( dx \) alongside \( x \) terms). For example, consider \( f' = 3f \), which can be rearranged to \( \frac{f'}{f} = 3 \), isolating \( f \) and \( x \) related terms.

• The next step is to integrate both sides independently; here, it yields \( \int \frac{1}{f} \, df = \int 3 \, dx \).
• This integration results in an expression involving \( \, ln|f| \) which can be solved using exponentiation.

The final solution is found by solving the integrated equations, and applying any given boundary conditions to find any constants involved. This technique is elegant because it allows complex equations to be broken down into simpler parts.

Characteristic Equations

Characteristic equations arise when solving higher order linear differential equations with constant coefficients. These equations are crucial for finding the general solutions of differential equations, especially those that are second-order or higher.
The process involves assuming a solution of the form \( f(x) = e^{rx} \) and determining the values of \( r \) that satisfy the differential equation. Consider the differential equation \( f'' + 2f' - 15f = 0 \). Substituting \( f(x) = e^{rx} \) leads to the characteristic equation \( r^2 + 2r - 15 = 0 \).

• This characteristic equation is a quadratic equation in terms of \( r \).
• Solving this quadratic gives roots that determine the general form of the solution.

Depending on whether the roots are real or complex, the solutions to the differential equation can take various forms.

• Real, distinct roots lead to exponential solutions.
• Complex roots lead to sinusoidal solutions due to Euler's formula.

Understanding characteristic equations allows you to transform complex differential equations into solvable Algebra problems, revealing much about the differential equation’s behavioral nature.