Question

Find the solution of the following boundary-value problems: (a) \(y^{\prime \prime}-y=-f(x), \quad y(0)=y(1)=0 .\) (b) \(y^{\prime \prime}-y=-f(x), \quad y^{\prime}(0)=y^{\prime}(1)=0 .\)

Short answer

In the given ODE \(y^{\prime \prime}-y=-f(x)\) with different boundary conditions: (a) \(y(0)=y(1)=0\) (b) \(y^{\prime}(0)=y^{\prime}(1)=0\) We found the general solution as \(y(x) = A e^x + Be^{-x} + u_1(x)e^x + u_2(x)e^{-x}\). To find the particular solutions for (a) and (b), we need to apply their respective boundary conditions to this general solution and solve for the constants \(A\), \(B\), \(u_1\), and \(u_2\).

Step-by-step solution

Find the complementary function for the homogenous equation

Solve the homogenous equation \(y^{\prime \prime}-y=0\) by assuming a solution of the form \(y=e^{lx}\), where \(l\) is a constant. Plugging this ansatz into the equation gives us $$l^2 e^{lx}-e^{lx}=0 \implies e^{lx}(l^2-1) = 0$$ Since \(e^{lx}\) can't be zero, we have $$l^2-1=0,$$ which gives us the roots \(l=1\) and \(l=-1\). The complementary function of the homogenous equation is thus $$y_c(x) = A e^x + Be^{-x},$$ where \(A\) and \(B\) are constants.

Find the particular integral for the nonhomogenous equation

We need to find a particular solution for the original non-homogenous ODE. The method of variation of parameters can be used to find the particular integral. Let a particular integral be of the form: $$y_p(x)=u_1(x)e^x+u_2(x)e^{-x}$$ From \(y_p\), compute: - \(y_p^\prime = u_1^\prime e^x - u_2^\prime e^{-x}\) - \(y_p^{\prime \prime} = u_1^{\prime\prime}e^x + u_2^{\prime\prime}e^{-x}\) Plug them into the non-homogeneous ODE: \(y_p^{\prime \prime} - y_p = -f(x)\) This gives: \(u_1^{\prime\prime}e^x + u_2^{\prime\prime}e^{-x} - (u_1(x)e^x+u_2(x)e^{-x}) = -f(x)\) Now we use the variation of parameters method and set: - \(u_1^\prime e^x - u_2^\prime e^{-x} = 0\) - \(u_1^{\prime\prime}e^x + u_2^{\prime\prime}e^{-x} = -f(x)\) From these two equations, solve for \(u_1^{\prime\prime}\), \(u_2^{\prime\prime}\), and integrate them twice to find \(u_1\) and \(u_2\). Then, substitute \(u_1\) and \(u_2\) to find the particular integral \(y_p(x)\). Now, we have the general solution: $$y(x) = y_c(x) + y_p(x) = A e^x + Be^{-x} + u_1(x)e^x + u_2(x)e^{-x}$$

Apply boundary conditions and solve for constants

For parts (a) and (b), we have different boundary conditions. (a) For \(y(0)=y(1)=0\), plug in these values to the general solution and solve for \(A\), \(B\), \(u_1\), and \(u_2\). This will give you the particular solution for part (a). (b) For \(y^{\prime}(0)=y^{\prime}(1)=0\), take the derivative of the general solution with respect to \(x\) and evaluate at the given points. This will give 2 equations to solve for the constants \(A\), \(B\), \(u_1\), and \(u_2\). Find the constants and then substitute them into the general solution to find the particular solution for part (b).

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving derivatives of a function with respect to one independent variable. They are formulated to model various phenomena such as motion, heat conduction, and wave propagation.

ODEs can be classified based on their order, with the order being the highest derivative present in the equation. The general form of an ODE is expressed as
\[ F(x, y, y', y'', ...y^{(n)}) = 0, \]where \(y^{(n)}\) denotes the \(n\)th derivative of \(y\) with respect to \(x\). In our exercise, we deal with a second-order ODE because the highest derivative is \(y''\), which is the second derivative of \(y\) with respect to \(x\).

Boundary-value problems for ODEs involve finding a solution that satisfies the differential equation as well as specific values at the boundaries of the domain, like the given scenarios where \(y(0)\) and \(y(1)\), or their derivatives, are defined.

Complementary function

The complementary function, often denoted as \(y_c\), is a solution to the homogeneous part of a differential equation. Homogeneous ODEs are those with no independent term, written as \(L[y] = 0\), where \(L\) is a differential operator.

For example, in a second-order linear homogeneous ODE, the complementary function is found by solving the equation \(y'' - p(x)y' - q(x)y = 0\). We look for solutions that form a basis for the solution space and are thus linearly independent. In the step-by-step solution, the complementary function to \(y'' - y = 0\) was found by assuming a solution of the form \(y = e^{lx}\) and solving for \(l\). The result was \(y_c(x) = A e^x + B e^{-x}\), where \(A\) and \(B\) are constants to be determined.

Particular integral

The particular integral, usually represented as \(y_p\), is the solution to the non-homogeneous part of the differential equation, which includes the independent term, in our case \(-f(x)\).

This solution addresses the specific input or forcing function of the ODE. The aim is not to solve the general ODE, but to find any single solution that satisfies the equation.

• If the ODE has constant coefficients, the method of undetermined coefficients can be quite effective.
• For ODEs with variable coefficients or for more complex forcing functions, the method of variation of parameters is a versatile tool.

This particular solution, when added to the complementary function, gives the general solution to the non-homogeneous ODE.

Method of variation of parameters

The method of variation of parameters is a technique used to find the particular integral of a non-homogeneous differential equation. It is especially useful when the non-homogeneous term \(-f(x)\) makes methods like undetermined coefficients less practical.

In this method, we assume that the coefficients \(A\) and \(B\) in the complementary function are actually functions of \(x\). That is, we seek a solution of the form \(y_p(x)=u_1(x)e^x+u_2(x)e^{-x}\). We then differentiate this assumed solution and match it up with our original ODE to create a system of equations.

• We calculate first and second derivatives of our assumed solution.
• We set up a system of equations that allows us to find \(u_1'\) and \(u_2'\) such that their combination satisfies the non-homogeneous part of the ODE.

Following this, we integrate to get expressions for \(u_1(x)\) and \(u_2(x)\), which are then substituted back to give us the particular integral \(y_p(x)\). The general solution to the ODE is the sum of the complementary function and the particular integral.