Question

State whether the given boundary value problem is homogeneous or non homogeneous. $$ -y^{\prime \prime}+x^{2} y=\lambda y, \quad y^{\prime}(0)-y(0)=0, \quad y^{\prime}(1)+y(1)=0 $$

Short answer

Differential equation: \(- y^{\prime \prime} + x^{2}y = \lambda y\) Boundary conditions: \(y^{\prime}(0) - y(0)=0\) and \(y^{\prime}(1) + y(1)=0\) Answer: The given boundary value problem is homogeneous.

Step-by-step solution

Analyze the differential equation

The given differential equation is: $$ - y^{\prime \prime} + x^{2}y = \lambda y $$ This is a linear differential equation. To determine if it's homogeneous or not, we need to examine the right-hand side of the equation, which is \(\lambda y\). Since \(\lambda y\) does not involve any function other than \(y\), this equation is homogeneous.

Analyze the boundary conditions

Now let's examine the boundary conditions: 1. \(y^{\prime}(0) - y(0)=0\) 2. \(y^{\prime}(1) + y(1)=0\) For boundary condition 1: $$ y^{\prime}(0) - y(0) = 0 \implies y^{\prime}(0) = y(0) $$ Since this equation is true when \(y^{\prime}(0)=0\) and \(y(0)=0\), the first boundary condition is homogeneous. For boundary condition 2: $$ y^{\prime}(1) + y(1) = 0 \implies y^{\prime}(1) = -y(1) $$ Similar to the first boundary condition, this equation is true when \(y^{\prime}(1)=0\) and \(y(1)=0\). Therefore, the second boundary condition is also homogeneous.

Determine if the boundary value problem is homogeneous or non-homogeneous

Since both the differential equation and the boundary conditions are homogeneous, the given boundary value problem is homogeneous.