Question

Solve the following initial value problems. $$u^{\prime \prime}(x)=4 e^{2 x}-8 e^{-2 x}, u(0)=1, u^{\prime}(0)=3$$

Short answer

Answer: The particular solution to the given IVP is: $$u(x) = -3x + 1 + e^{2x} - 2e^{-2x}$$

Step-by-step solution

Identify the structure of the given ODE

The given ODE is $$u''(x) = 4 e^{2x} - 8 e^{-2x}$$ This is a second-order linear inhomogeneous ODE, with the inhomogeneous term being the combination of two exponentials with different exponents.

Find the general solution of the homogeneous ODE

The homogeneous part of the ODE is $$u''(x) = 0$$ The solution to this ODE is given by the sum of a linear function and a constant: $$u_h(x) = Ax + B$$

Find a particular solution to the inhomogeneous ODE

Because the inhomogeneous term has the form of a combination of two exponentials, we will make an ansatz for u_p(x) in the same form: $$u_p(x) = C e^{2x} + D e^{-2x}$$ Now, taking the first and second derivatives of u_p(x), we have: $$u_p'(x) = 2C e^{2x} - 2D e^{-2x}$$ and $$u_p''(x) = 4C e^{2x} + 4D e^{-2x}$$ Substituting these expressions into the given ODE, we obtain: $$4C e^{2x} + 4D e^{-2x} = 4 e^{2x} - 8 e^{-2x}$$ Comparing the coefficients of the exponentials, we get: - For \(e^{2x}\): \(4C = 4\), which gives us \(C = 1\). - For \(e^{-2x}\): \(4D = -8\), which gives us \(D=-2\). Thus, the particular solution is: $$u_p(x) = e^{2x} - 2e^{-2x}$$

Combine solutions

The general solution to the inhomogeneous ODE is the sum of the homogeneous and particular solutions: $$u(x) = Ax + B + e^{2x} - 2e^{-2x}$$

Apply initial conditions

To find the particular solution corresponding to this IVP, we apply the initial conditions: 1. \(u(0) = 1\): $$A(0) + B + e^{2(0)} - 2e^{-2(0)} = 1$$ Substituting the given values, we get: $$B = 1\} 2. \(u'(0) = 3\): $$u'(x) = A + 2e^{2x} + 4e^{-2x}$$ Substituting the initial condition \(u'(0) = 3\), we get: $$A + 2e^{2(0)} + 4e^{-2(0)} = 3$$ Solving for A, we get: $$A = 3 - 2 - 4 = -3$$ Thus, the particular solution to the IVP is: $$u(x) = -3x + 1 + e^{2x} - 2e^{-2x}$$

Key concepts

Initial Value Problems

An initial value problem (IVP) is a type of differential equation along with specific conditions that are provided at the start of the interval of interest, usually when the input variable is zero. The goal is to find a function that satisfies both the differential equation and the initial conditions.

For example, suppose you are given a second-order differential equation and two initial conditions like in the exercise:

• The differential equation: \( u''(x) = 4e^{2x} - 8e^{-2x} \)
• The initial conditions: \( u(0) = 1 \) and \( u'(0) = 3 \)

These initial conditions allow us to find a particular solution that passes through the specific point \( x = 0 \) with the given derivatives. Solving the IVP involves finding the general solution of the differential equation first and then applying these conditions to determine any unknown constants.

Linear Inhomogeneous ODE

A linear inhomogeneous ordinary differential equation (ODE) is an equation that includes a non-zero term, which is not a multiple of the unknown function and its derivatives. In practical terms, it means there is an extra function on the right side of the equation that makes it 'inhomogeneous' compared to a 'homogeneous' equation.

In the exercise, the ODE \( u''(x) = 4e^{2x} - 8e^{-2x} \) is inhomogeneous due to the presence of the terms \( 4e^{2x} \) and \( -8e^{-2x} \). These terms do not contain the unknown function \( u(x) \) itself, thus making this differential equation inhomogeneous.

To solve a linear inhomogeneous ODE, we often break the solution process into finding two parts: the solution to the corresponding homogeneous equation and a particular solution to the inhomogeneous equation.

Homogeneous Solutions

In solving differential equations, the term 'homogeneous solution' refers to the solution of the homogeneous version of the original ODE, which excludes any inhomogeneous (non-zero) terms.

For a homogeneous ODE, we assume all terms are multiples of the unknown function or its derivatives, like \( u''(x) = 0 \), which was isolated from the original exercise. Solving it yields a function where adding a constant or linear term completes its general solution.
In this case, solving \( u''(x) = 0 \) results in the simplest general solution:

• \( u_h(x) = Ax + B \)

This representation, with constants \( A \) and \( B \), provides the foundation to which we add a particular solution to form the complete solution for the original inhomogeneous problem.

Particular Solutions

The particular solution addresses the inhomogeneous component of an ODE and specifically satisfies the non-homogeneous part. It works in conjunction with the general homogeneous solution to solve the overall inhomogeneous ODE.

For the exercise at hand, the inhomogeneous equation involved \( 4e^{2x} - 8e^{-2x} \) as external inputs. By assuming a particular form similar to these inputs, we guessed a solution: \( u_p(x) = Ce^{2x} + De^{-2x} \). We then determine the constants \( C \) and \( D \) by substituting back into the modified ODE:

• Matching coefficients: For \( e^{2x} \), \( 4C = 4 \) thus \( C = 1 \)
• For \( e^{-2x} \), \( 4D = -8 \) thus \( D = -2 \)

Thus, the particular solution is \( u_p(x) = e^{2x} - 2e^{-2x} \). When combined with the homogeneous solution, we obtain the total solution that can be further refined using the initial conditions.