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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.
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:
  • 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.
Understanding how to work with differential equations opens doors to solving complex problems about how things evolve and interact.
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:
  • 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 \(y\).
Effective use of boundary conditions helps tailor solutions to real-world requirements, ensuring they are both accurate and applicable.

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Most popular questions from this chapter

Solve the given problem by means of an eigenfunction expansion. $$ y^{\prime \prime}+2 y=-x, \quad y(0)=0, \quad y(1)=0 $$

deal with column buckling problems. In some buckling problems the eigenvalue parameter appears in the boundary conditions as well as in the differential equation. One such case occurs when one end of the column is clamped and the other end is free. In this case the differential equation \(y^{i v}+\lambda y^{\prime \prime}=0\) must be solved subject to the boundary conditions $$ y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(L)=0, \quad y^{\prime \prime \prime}(L)+\lambda y^{\prime}(L)=0 $$ Find the smallest eigenvalue and the corresponding eigenfunction.

Use eigenfunction expansions to find the solution of the given boundary value problem. $$ \begin{array}{ll}{u_{t}=u_{x x}-x,} & {u(0, t)=0, \quad u_{x}(1, t)=0, \quad u(x, 0)=\sin (\pi x / 2)} \\ {\text { see Problem } 2}\end{array} $$

By a procedure similar to that in Problem 28 show that the solution of the boundary value problem $$ -\left(y^{\prime \prime}+y\right)=f(x), \quad y(0)=0, \quad y(1)=0 $$ is $$ y=\phi(x)=\int_{0}^{1} G(x, s) f(s) d s $$ where $$ G(x, s)=\left\\{\begin{array}{ll}{\frac{\sin s \sin (1-x)}{\sin 1},} & {0 \leq s \leq x} \\ {\frac{\sin x \sin (1-s)}{\sin 1},} & {x \leq s \leq 1}\end{array}\right. $$

Use the method of Problem 11 to transform the given equation into the form \(\left[p(x) y^{\prime}\right]'+q(x) y=0\) $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-v^{2}\right) y=0, \quad \text { Bessel equation } $$

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