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Magazine Chapter 10: Problem 5
Either solve the given boundary value problem or else show that it has no solution. \(y^{\prime \prime}+y=x, \quad y(0)=0, \quad y(\pi)=0\)
Short Answer
Expert verified
Answer: No, the given boundary value problem does not have a solution because the second boundary condition cannot be satisfied.
Step by step solution
01
Identify the given differential equation and boundary conditions
The given differential equation is:
\(y^{\prime \prime}+y=x\)
The boundary conditions are:
\(y(0)=0\) and \(y(\pi)=0\)
02
Convert the given differential equation into an initial value problem (IVP)
Solve the differential equation as a homogeneous equation first. Substitute \(x=0\) into the differential equation to get:
\(y^{\prime \prime}(0) + y(0) = 0\)
Since we know \(y(0)=0\), we can write the IVP as:
\(y^{\prime \prime}+y = 0; \quad y(0)=0, \quad y^{\prime}(0)=A\)
Where \(A\) is a constant that we will determine later.
03
Solve the homogeneous equation
The homogeneous equation is
\(y^{\prime \prime}+y=0\)
The auxiliary equation (or characteristic equation) is:
\(m^2+1=0\)
The roots of this equation are \(m= \pm i\). Now, we can write the general solution of this homogeneous equation as:
\(y_h(x) = C_1 \cos(x) + C_2 \sin(x)\)
04
Solve the nonhomogeneous equation
The nonhomogeneous equation can be written as:
\(y^{\prime \prime}+y=x\)
Applying the method of undetermined coefficients, the particular solution is in the form of: $$ y_p(x) = Ax + B $$
Differentiate \(y_p(x)\) twice w.r.t x to get:
$$y_p^{\prime}(x) = A, \quad y_p^{\prime \prime}(x) = 0$$
Substitute the above equations in the given nonhomogeneous equation to get: \((0) + (Ax + B) = x\)
Thus, $$A = 1$$ and $$B = 0$$ Ensuring $$y_p(x) = x$$ is the particular solution to the nonhomogeneous differential equation.
05
Write the general solution of the nonhomogeneous equation
The general solution of the nonhomogeneous equation is the sum of the solutions of the homogeneous and nonhomogeneous equation. Therefore,
$$y(x) = y_h(x) + y_p(x) = C_1 \cos(x) + C_2 \sin(x) + x$$
06
Apply the boundary conditions to find the coefficients
Apply the first boundary condition \(y(0) = 0\):
$$y(0) = C_1\cos(0) + C_2\sin(0) + 0 = 0$$
$$C_1 = 0$$
Now apply the second boundary condition \(y(\pi) = 0\):
$$y(\pi) = C_1 \cos(\pi) + C_2 \sin(\pi) + \pi = 0$$
$$0 + 0 + \pi = 0$$
But that's not possible, so the BVP does not have a solution.
07
Conclusion
Since we cannot satisfy the second boundary condition, the given boundary value problem has no solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are a type of mathematical equation that involve derivatives鈥攅xpressions that indicate the rate of change of a function with respect to one or more variables. They play a crucial role in modeling a wide range of phenomena in engineering, physics, biology, economics, and beyond. By studying how things change, we can predict future behavior or the outcome of various processes.
When it comes to solving these equations, there are two main types to consider: homogeneous and nonhomogeneous equations. A homogeneous differential equation is characterized by the absence of a standalone function term, like the example \(y'' + y = 0\), while a nonhomogeneous differential equation includes an added function, such as \(y'' + y = x\) in our exercise. The first step in solving these is to determine the type and order of the equation, then apply the appropriate method to find the solution.
When it comes to solving these equations, there are two main types to consider: homogeneous and nonhomogeneous equations. A homogeneous differential equation is characterized by the absence of a standalone function term, like the example \(y'' + y = 0\), while a nonhomogeneous differential equation includes an added function, such as \(y'' + y = x\) in our exercise. The first step in solving these is to determine the type and order of the equation, then apply the appropriate method to find the solution.
Homogeneous Equations
Homogeneous equations are simpler forms of differential equations in which all terms depend on the unknown function and its derivatives. They do not contain a term that is simply a function of the independent variable鈥攍ike \(x\) in our nonhomogeneous example. In the equation \(y'' + y = 0\), for instance, the right side equals zero, marking it as homogeneous.
To solve such an equation, we typically look for solutions that can be expressed as sums of simple functions of the independent variable, like sines and cosines. For example, if the characteristic equation of the homogeneous part is \(m^2+1=0\), then its roots will be imaginary numbers \(m=\pm i\), and the solution would be a combination of \(\cos(x)\) and \(\sin(x)\) functions. Knowing how to solve these foundational equations is critical for addressing more complex nonhomogeneous cases.
To solve such an equation, we typically look for solutions that can be expressed as sums of simple functions of the independent variable, like sines and cosines. For example, if the characteristic equation of the homogeneous part is \(m^2+1=0\), then its roots will be imaginary numbers \(m=\pm i\), and the solution would be a combination of \(\cos(x)\) and \(\sin(x)\) functions. Knowing how to solve these foundational equations is critical for addressing more complex nonhomogeneous cases.
Method of Undetermined Coefficients
The method of undetermined coefficients is a technique for finding particular solutions to nonhomogeneous linear differential equations. It's most effective when the nonhomogeneous term鈥攖he standalone function on one side of the equation鈥攊s a simple function, such as a polynomial, exponential, sine, or cosine.
To apply the method, you first guess a form for the particular solution. The 'undetermined coefficients' in this form are then found by substituting the guess back into the original equation. For example, in our problem \(y''+y=x\), we guessed that the particular solution would look like \(y_p(x) = Ax + B\), where \(A\) and \(B\) are constants to be found. Differentiating our guess for \(y_p\) and substituting it back into the equation allows us to find the values for these constants.
This method is an excellent shortcut when the right type of nonhomogeneous term is involved, as it bypasses more cumbersome general solution methods that might not provide such direct insights into the structure of the solution.
To apply the method, you first guess a form for the particular solution. The 'undetermined coefficients' in this form are then found by substituting the guess back into the original equation. For example, in our problem \(y''+y=x\), we guessed that the particular solution would look like \(y_p(x) = Ax + B\), where \(A\) and \(B\) are constants to be found. Differentiating our guess for \(y_p\) and substituting it back into the equation allows us to find the values for these constants.
This method is an excellent shortcut when the right type of nonhomogeneous term is involved, as it bypasses more cumbersome general solution methods that might not provide such direct insights into the structure of the solution.
