Question

Solve the following boundary value problems directly, when possible. a. \(x^{\prime \prime}+x=2, \quad x(0)=0, \quad x^{\prime}(1)=0 .\) b. \(y^{\prime \prime}+2 y^{\prime}-8 y=0, \quad y(0)=1, \quad y(1)=0\). c. \(y^{\prime \prime}+y=0, \quad y(0)=1, \quad y(\pi)=0\).

Short answer

The general solutions for the given differential equations are: (a) \(x(t)=-2cos(t)+2sin(t)+2\), (b) \(y(t)=\frac{1}{3}e^{2t}+\frac{2}{3}e^{-4t}\), and (c) \(y(t)=cos(t)\)

Step-by-step solution

Solve Equation (a)

The equation \(x^{\prime \prime}+x=2\) is a non-homogeneous differential equation. The general solution is formed of a homogeneous solution and a particular solution. From characteristic equation \(m^2+1=0\), we get \(m=\pm i\). Hence, the homogeneous solution is \(x_h(t)=C_1cos(t)+C_2sin(t)\). The particular solution can be guessed as \(x_p=a\). Subsituting \(x_p\) into the non-homogeneous differential equation yields \(a=2\). So, the general solution is \(x(t)=C_1cos(t)+C_2sin(t)+2\). Apply the initial conditions \(x(0)=0\) gives \(C_1= -2\). Also, \(x^{\prime}(1)=0\), gives \(C_2=2\). Therefore, the solution is \(x(t)=-2cos(t)+2sin(t)+2\)

Solve Equation (b)

The equation \(y^{\prime \prime}+2 y^{\prime}-8 y=0\) is a homogeneous differential equation. From the characteristic equation \(m^2+2m-8=0\), we find \(m_1=2, m_2=-4\). Hence, the general solution is \(y(t)=C_1e^{2t}+C_2e^{-4t}\). Applying the boundary conditions gives us \(C_1+C_2=1\) for \(y(0)=1\) and \(C_1e^{2}+C_2e^{-4}=0\) for \(y(1)=0\). Solving these simultaneous equations yields \(C_1=1/3\) and \(C_2=2/3\). Therefore, the solution to the differential equation is \(y(t)=\frac{1}{3}e^{2t}+\frac{2}{3}e^{-4t}\)

Solve Equation (c)

The equation \(y^{\prime \prime}+y=0\) is a homogeneous differential equation. From the characteristic equation \(m^2+1=0\), we get \(m=\pm i\). Hence, the general solution is \(y(t)=C_1cos(t)+C_2sin(t)\). Apply the initial conditions \(y(0)=1\) gives \(C_1= 1\). Also, \(y(\pi)=0\), gives \(C_2=0\). Therefore, the solution is \(y(t)=cos(t)\)

Key concepts

Understanding Non-Homogeneous Differential Equations

A non-homogeneous differential equation is one that includes a term independent of the variable function and its derivatives, known as the non-homogeneous term. In essence, it consists of a homogeneous part, which can be written without any external inputs, and a non-homogeneous part where external forces or sources are considered.

To solve a non-homogeneous equation, you generally find two parts:

• Homogeneous Solution (\(x_h\)): Solve the respective homogeneous differential equation, which is formed by setting the non-homogeneous part to zero.
• Particular Solution (\(x_p\)): Specifically tailor a solution that satisfies the entire equation, including the non-homogeneous term.

Combine these to form the general solution. For example, in the equation from the problem \(x'' + x = 2\), the particular solution is consistent with \(x_p = 2\). Together, with the homogeneous solution, forms the complete solution.

Exploring Homogeneous Differential Equations

Homogeneous differential equations are framed as equal to zero. The term homogeneous indicates that the flow or input equals zero outside the system. Such equations rely entirely on their intrinsic features without any external influence.

The general form is often represented as \(a_0 x^{(n)} + a_1 x^{(n-1)} + ... + a_n x = 0\).

They are solved by finding the characteristic roots that formulate the general solution as a linear combination of functions dictated by those roots.

For example, in our problem, equation (b) \(y'' + 2y' - 8y = 0\) is homogeneous. The general solution established from the characteristic equation gave two distinct exponential terms illustrating the system's natural response.

The Characteristic Equation in Differential Equations

The characteristic equation is a valuable step in solving both homogeneous and linear non-homogeneous differential equations. It is derived by replacing the derivatives in the differential equation with powers of a variable, often denoted as \(m\).

For example, suppose we have the differential equation \(ay'' + by' + cy = 0\). By substituting \(y = e^{mt}\) into the equation, we transform it into a polynomial equation \(am^2 + bm + c = 0\). This polynomial is the characteristic equation. Solving it gives us roots (\(m_1, m_2, ...\)), which in turn informs the form of the general solution.

Depending on whether the roots are real, complex, or repeated, they primarily dictate the nature of the solution whether as exponential functions, trigonometric functions, or their combinations. This method was utilized in the example problems where solutions of both real and complex roots led to distinctive forms of solutions.