Chapter 11: Problem 5
State whether the given boundary value problem is homogeneous or non homogeneous. $$ -\left[\left(1+x^{2}\right) y^{\prime}\right]=\lambda y+1, \quad y(-1)=0, \quad y(1)=0 $$
Short Answer
Expert verified
**Answer:** The given boundary value problem is nonhomogeneous.
Step by step solution
01
Identify the Differential Equation and Boundary Conditions
Examine the given BVP. The differential equation is $$-[(1+x^2)y'] = \lambda y + 1$$ and the boundary conditions are $$y(-1) = 0$$ and $$y(1) = 0$$.
02
Check if the Right-Hand Side of the Differential Equation is Zero
In this problem, the right-hand side of the differential equation is given by: $$\lambda y + 1$$, we need to check if this is identically equal to zero.
03
Determine Whether the BVP is Homogeneous or Nonhomogeneous
Since the right-hand side of the differential equation is not zero, it means that the given boundary value problem is nonhomogeneous.
The given boundary value problem is: $$-\left[\left(1+x^{2}\right) y^{\prime}\right]=\lambda y+1, \quad y(-1)=0, \quad y(1)=0$$ It is a **nonhomogeneous** BVP because the right-hand side of the differential equation is not identically zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous vs Nonhomogeneous Equations
When dealing with boundary value problems (BVPs), it's important to classify them as either homogeneous or nonhomogeneous. This classification helps in understanding how solutions can be approached and solved.
A homogeneous equation is one in which every term is a function of the dependent variable and possibly its derivatives, without a standalone constant term. Mathematically, if we look at an equation like \(-[(1+x^2)y'] = \lambda y\), it's homogeneous as there is no extra term adding to the equation.
Nonhomogeneous equations, on the other hand, contain additional non-zero terms, often constants or functions of the independent variable. In our boundary value problem, the equation \(-[(1+x^2)y'] = \lambda y + 1\) includes a \(+1\), which makes it nonhomogeneous. This extra term indicates that external factors or sources are influencing the system. Identifying this difference is crucial because the methods used to solve homogeneous equations generally differ from those used for nonhomogeneous ones.
A homogeneous equation is one in which every term is a function of the dependent variable and possibly its derivatives, without a standalone constant term. Mathematically, if we look at an equation like \(-[(1+x^2)y'] = \lambda y\), it's homogeneous as there is no extra term adding to the equation.
Nonhomogeneous equations, on the other hand, contain additional non-zero terms, often constants or functions of the independent variable. In our boundary value problem, the equation \(-[(1+x^2)y'] = \lambda y + 1\) includes a \(+1\), which makes it nonhomogeneous. This extra term indicates that external factors or sources are influencing the system. Identifying this difference is crucial because the methods used to solve homogeneous equations generally differ from those used for nonhomogeneous ones.
Differential Equations
Differential equations are equations involving the derivatives of a function. They describe how a particular quantity changes over time or space and are essential in modeling real-world phenomena across physics, engineering, biology, and beyond.
In our boundary value problem, the given differential equation is \(-[(1+x^2)y'] = \lambda y + 1\). Here, \(y'\) represents the derivative of \(y\) with respect to \(x\), showing how \(y\) changes as \(x\) varies.
Here are some key points to understand about differential equations:
In our boundary value problem, the given differential equation is \(-[(1+x^2)y'] = \lambda y + 1\). Here, \(y'\) represents the derivative of \(y\) with respect to \(x\), showing how \(y\) changes as \(x\) varies.
Here are some key points to understand about differential equations:
- Order: The highest derivative present determines the order. In our case, this is a first-order differential equation.
- Linear vs Nonlinear: It's linear if the dependent variable and its derivatives appear to the first power and are not multiplied together. Our example is linear due to this form.
- Applications: They model changes in systems, like motion, heat, wave propagation, and more.
Boundary Conditions
Boundary conditions are additional constraints needed to find a unique solution to a boundary value problem. They specify the values of a solution at particular points, which is essential since many differential equations can have numerous or infinite solutions.
For the boundary value problem we're looking at, the boundary conditions are \(y(-1) = 0\) and \(y(1) = 0\). These conditions specify that the function \(y\) must be zero at \(x = -1\) and \(x = 1\), limiting the possible solutions to those that satisfy these points.
Here's why boundary conditions are vital:
For the boundary value problem we're looking at, the boundary conditions are \(y(-1) = 0\) and \(y(1) = 0\). These conditions specify that the function \(y\) must be zero at \(x = -1\) and \(x = 1\), limiting the possible solutions to those that satisfy these points.
Here's why boundary conditions are vital:
- Uniqueness: Proper boundary conditions ensure there is a unique solution, critical for practical applications.
- Physical Meaning: They often represent actual constraints in physical problems, like temperature or velocity at a surface.
- Solution Approach: Different methods of solving differential equations apply depending on the type of boundary conditions, such as Dirichlet, Neumann, or Robin conditions. In our problem, they're Dirichlet conditions since they specify values of \(y\).