Question

Use the method of eigenfunction expansions to solve the problems: a. \(y^{\prime \prime}+4 y=x^{2}, \quad y^{\prime}(0)=y^{\prime}(1)=0\). b. \(y+4 y=x^{2} \quad y(0)=y(1)=0\).

Short answer

The solution to part (a) is \( y = (1/4) x^{2} \) and the solution to part (b) is \( y = x - x^{3}/6 \)

Step-by-step solution

Solve the Homogeneous Differential Equation

First, let's solve the homogeneous equations for both problems. Problem a: \( y'' + 4y = 0 \), and problem b: \( -4y = 0 \). Both of these equations are solved by assuming the solution is of the form \( y = e^{rx} \), where \( r \) is root of the characteristic equation associated with the differential equation. For problem a, the characteristic equation is \( r^{2} + 4 = 0 \). The solutions are \( r = ±2i \). For problem b, the characteristic equation is actually a linear equation (as there is no derivative term), giving \( r = 0 \) as its only solution.

Construct Eigenfunction Expansions

Using the roots from step 1, we form the eigenfunction expansions. For problem a, the general solution of the homogeneous differential equation is \( y(x) = C_{1} cos(2x) + C_{2} sin(2x) \). We should impose the condition \( y'(0)= y'(1) = 0 \) on this solution. The derivative of y is \( y'(x) = -C_{1} 2sin(2x) + C_{2} 2cos(2x) \). Imposing the boundary conditions, we solve for the constants to get \( C_{1}=0 \) and \( C_{2}=0 \) which makes the solution trivial (i.e., y = 0) for all x. For problem b, since the only root is \( r=0 \), the eigenfunction is a straight line \( y(x)=C \). Imposing the homogeneous boundary conditions \( y(0)=y(1)=0 \), we solve for constants to find again, \( C = 0 \), which is the only solution satisfying both of them.

Integrate the inhomogeneous part

Finally, we add the particular solution to the inhomogeneous equations. Problem a: The inhomogeneous equation is \( y'' + 4y = x^{2} \). The particular solution of this equation is given by a quadratic polynomial in \( x \), say \( y(x) = Ax^{2} + Bx + C \). But the boundary conditions dictate that \( B = 0 \) and will make \( C = 0 \) as well. When we solve for \( A \) we get that \( y(x) = (1/4) x^{2} \). For problem b: the inhomogeneous part is a linear term \( y = x^{2} \). The particular solution of the inhomogeneous equation is \( y(x) = Ax^{2}+ Bx + C \). After solving for A, B, and C under the boundary conditions we find \( y(x) = x - x^{3}/6 \). Thus the solutions to both problems are: a.) \( (1/4) x^{2} \) b.) \( x - x^{3}/6 \)

Key concepts

Homogeneous Differential Equation

Understanding a homogeneous differential equation is critical for tackling complex mathematical problems. Such an equation is characterized by being set equal to zero and typically depicts a relationship involving derivatives of a function but does not include any terms without derivatives. For instance, in our example problem a: \( y'' + 4y = 0 \), only derivatives of \( y \) and \( y \) itself appear in the equation, and there are no external forces or inputs that alter the function's behavior—inherent properties of the system determine its evolution.

One typically solves homogeneous differential equations by proposing a specific type of solution, often an exponential function of the form \( y = e^{rx} \) where \( r \) is a constant that must satisfy the characteristic equation derived from the differential equation. This approach turns a differential equation into an algebraic one, making it more tractable. The solution to such equations has great importance as it provides the complementary or homogeneous solution, to which we add particular solutions when dealing with the inhomogeneous counterpart.

Characteristic Equation

The characteristic equation plays a pivotal role in solving homogeneous differential equations. By substituting an assumed solution form into the differential equation, one can derive this algebraic equation. For example, assuming \( y = e^{rx} \) for problem a and substituting it into the equation gives us \( r^{2} e^{rx} + 4e^{rx} = 0 \), which simplifies to the characteristic equation \( r^{2} + 4 = 0 \). This equation is crucial because its roots dictate the nature of the solution to the differential equation. Complex roots, as seen with \( ±2i \) in problem a, indicate oscillatory solutions, whereas real roots would suggest exponentially growing or decaying solutions. It is by scrutinizing the characteristic equation that we can anticipate the system's behavior without solving the entirety of the equation.

Boundary Conditions

To solve differential equations fully, one must take into account the boundary conditions. These conditions provide additional information about the values or derivatives of the solution at specific points, which is necessary to determine the unknown constants within the general solution. In the cases studied, we are given conditions for the derivative of \( y \) at \( x = 0 \) and \( x = 1 \) for problem a.

By applying these boundary conditions, we can solidify our solution: when we set the derivative of our general solution equal to these boundary values, we can solve for the constants involved. However, as seen in our examples, sometimes the boundary conditions can lead to trivial solutions, such as all constants being zero, indicating that the homogeneous solution does not contribute to the final form of the system.

Particular Solution

Adding to our toolbox of solutions, the particular solution is specific to inhomogeneous differential equations. It is the part of the general solution that directly balances the non-zero side of the inhomogeneous equation. Typically, the form of the particular solution is 'guessed' based on the non-homogeneous term and then precisely determined by substituting it into the differential equation.

For problem a's inhomogeneous part, \( x^2 \), a polynomial of the same degree is assumed for the particular solution. This gives us a quadratic polynomial \( Ax^2 + Bx + C \). After identifying the coefficients that satisfy the inhomogeneous differential equation given the boundary conditions, the particular solution for problem a becomes \( (1/4) x^2 \) which reflects the forcing term without the restrictions of the homogeneous equation's solution.

Inhomogeneous Differential Equation

An inhomogeneous differential equation is more complicated than its homogeneous counterpart because it includes a non-zero term representing an external input or driving force, like \( x^2 \) in problem a: \( y'' + 4y = x^2 \). Solving these equations involves finding both the complementary (or homogeneous) solution and a particular solution. The latter must be added to account for the inhomogeneous term.

The approach is to first determine the general solution to the associated homogeneous equation, which does not include the inhomogeneous term. Then, a particular solution that directly addresses the inhomogeneous term is proposed and found. For a complete solution, one combines the homogeneous solution (adjusted by the boundary conditions) and the particular solution. In problem b, once the particular solution \( y(x) = x - x^{3}/6 \) that satisfies the boundary conditions is found, it represents the whole solution since the homogeneous solution contributes trivially in this case.