Question

Either solve the given boundary value problem or else show that it has no solution. \(y^{\prime \prime}+4 y=\sin x, \quad y(0)=0, \quad y(\pi)=0\)

Short answer

Question: Determine if the boundary value problem \(y^{\prime \prime}+4y = \sin{x}\), given the boundary conditions \(y(0) = 0\) and \(y(\pi) = 0\), has a solution, and find the solution if it exists. Solution: We followed the steps of finding a complementary solution and a particular solution and then combined them to find the general solution. After applying the given boundary conditions, we found a valid solution \(y(x) = \frac{1}{3}\sin(x)\) for the boundary value problem.

Step-by-step solution

Find the complementary solution

To find the complementary solution (i.e, the solution without the non-homogeneous term), we will solve the ODE with the non-homogeneous term set to zero. \(y^{\prime \prime}+4 y=0\) This is a second order linear ODE with constant coefficients, so we can solve using the characteristic equation: \(r^2 + 4 = 0\) ⟹ \(r = \pm 2i\) Since we have complex roots, the complementary solution can be represented as: \(y_c(x) = A\cos(2x) + B\sin(2x)\)

Find the particular solution

To find a particular solution for the given non-homogeneous equation, we can use the method of undetermined coefficients. Since the non-homogeneous term is a sinusoidal function (\(\sin x\)), we can assume the particular solution to be in the form \(y_p(x) = C\cos(x) + D\sin(x)\) Now we need to find the first and second derivative of \(y_p(x)\): \(y_p^\prime(x) = -C\sin(x) + D\cos(x)\) \(y_p^{\prime \prime}(x) = -C\cos(x) - D\sin(x)\) Substitute the expressions for \(y_p(x)\), \(y_p^\prime(x)\), and \(y_p^{\prime \prime}(x)\) into the given ODE: \((-C\cos(x) - D\sin(x)) + 4(C\cos(x) + D\sin(x)) = \sin(x)\) Now, equating coefficients of \(\sin(x)\) and \(\cos(x)\), we have: \((-D + 4D) = 1,\) ⟹ \(D = \frac{1}{3}\) \((-C + 4C) = 0,\) ⟹ \(C = 0\) Thus, the particular solution is: \(y_p(x) = \frac{1}{3}\sin(x)\)

Find the general solution

Now we can combine the complementary and particular solutions to find the general solution: \(y(x) = y_c(x) + y_p(x)\) ⟹ \(y(x) = A\cos(2x) + B\sin(2x) + \frac{1}{3}\sin(x)\)

Apply the boundary conditions

Now let's apply the given boundary conditions one by one. Boundary condition 1: \(y(0) = 0\) ⟹ \(0 = A\cos(0) + B\sin(0) + \frac{1}{3}\sin(0)\) ⟹ \(A = 0\) Boundary condition 2: \(y(\pi) = 0\) 0 = 0 + B\sin(2\pi) + \frac{1}{3}\sin(\pi)$ ⟹ \(B = 0\)

Check if the exercise has a solution

Since applying the given boundary conditions yields a trivial solution (\(A = 0\) and \(B = 0\)), it means that the exercise indeed has a solution: \(y(x) = \frac{1}{3}\sin(x)\) Thus, the given boundary value problem has a valid solution.

Key concepts

Complementary Solution

When solving a boundary value problem that involves a differential equation, one of the first steps is to find the complementary solution, often denoted as y_c(x). The complementary solution corresponds to the general solution of the homogeneous version of the equation, which means the non-homogeneous term (such as a function of x on the right-hand side of the equation) is set to zero. In the context of the given exercise, the homogeneous equation is y'' + 4y = 0.

This type of equation is a linear differential equation with constant coefficients, which can be solved by finding the roots of the characteristic equation. In the provided solution, after determining the roots to be complex numbers, the complementary solution takes the form of a combination of sine and cosine functions: Acos(2x) + Bsin(2x). The constants A and B are to be determined using boundary conditions. Exactly this step reflects the intrinsically oscillatory nature of the solution to a linear second-order differential equation with complex roots.

Particular Solution

The particular solution, denoted as y_p(x), specifically addresses the non-homogeneous part of the differential equation. Unlike the complementary solution that reflects the general behavior of the system, the particular solution represents a single, specific function that satisfies the non-homogeneous equation. For the given problem, y'' + 4y = sin x, the particular solution is determined by considering the form of the non-homogeneous term, in this case, sin x.

The method of undetermined coefficients comes into play for finding this particular solution. It involves assuming a form for y_p(x) that is similar to the non-homogeneous part but with undetermined coefficients, which are later found by substituting y_p(x) into the original differential equation and equating coefficients. For a sinusoidal non-homogeneous term, the guess for the particular solution would typically be a linear combination of sine and cosine functions, which in this case is found to be (1/3)sin x after determining the coefficients.

Method of Undetermined Coefficients

The method of undetermined coefficients is a common technique for finding a particular solution to a non-homogeneous linear differential equation. The key idea is to propose a form for the particular solution with unknown coefficients and then determine these coefficients by substituting the proposed solution back into the differential equation.

This method works well when the non-homogeneous term is a simple function, such as polynomials, exponentials, or sine and cosine functions. In our exercise, because the non-homogeneous term is sin x, we assume a particular solution of the form Ccos x + Dsin x. After differentiation and substitution, the coefficients C and D are found by comparing both sides of the equation, resulting in a calculated particular solution. It's crucial to ensure that the form of the assumed particular solution does not overlap with the complementary solution. If it does, modification by multiplying with x is necessary to avoid redundancy and to correctly identify the undetermined coefficients.

Linear Differential Equation

Linear differential equations are equations that involve an unknown function and its derivatives. They are linear in terms of the function and its derivatives, without any products of the function and its derivatives or any powers other than the first. The general form of a second-order linear differential equation is ay'' + by' + cy = f(x), where a, b, and c are constants, and f(x) represents the non-homogeneous part of the equation.

In the context of the boundary value problem given, y'' + 4y = sin x is a second-order linear differential equation with the non-homogeneous term sin x. Such an equation is characterized by its principle of superposition, which allows the general solution to be expressed as a sum of the complementary and the particular solutions. Ultimately, to fully solve the boundary value problem, one must apply the boundary conditions to this general solution, which then determines any constants involved, resulting in a specific solution best customized to the initial problem set forth.