Let λ ∈ R be a parameter. We consider the ODE with conditions to the limits. y + λy = 0, y(0)= 0, y(1)=0 Explain why the existence and uniqueness theorem does not apply.

Name: Let λ ∈ R be a parameter. We consider the ODE with conditions to the limits. y + λy = 0, y(0)= 0, y(1)=0 Explain why the existence and uniqueness theorem does not apply.
Uploaded: 2022-02-23T17:22:57+00:00
Description: Let λ ∈ R be a parameter. We consider the ODE with conditions to the limits. y + λy = 0, y(0)= 0, y(1)=0 Explain why the existence and uniqueness theorem does not apply.

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Let λ ∈ R be a parameter. We consider the ODE with conditions to the limits.    y + λy = 0, y(0)= 0, y(1)=0     Explain why the existence and uniqueness theorem does not apply.

nick1337 · Accepted Answer

The given ODE is&amp;nbsp;y+λy=0, with initial conditions&amp;nbsp;y(0)=0&amp;nbsp;and&amp;nbsp;y(1)=0.To apply the existence and uniqueness theorem for ODEs, the equation must be in the form of&amp;nbsp;y&#039;=f(x,y)&amp;nbsp;with continuous partial derivatives of&amp;nbsp;f(x,&amp;nbsp;y)&amp;nbsp;with respect to y in some rectangular region containing the initial point.In this case, we have&amp;nbsp;y&#039;+λy=0, which can be written as&amp;nbsp;y&#039;=-λy. However, the function&amp;nbsp;f(x,y)=-λy&amp;nbsp;does not have continuous partial derivatives with respect to y in the entire xy-plane. Therefore, the existence and uniqueness theorem does not apply to this ODE.Moreover, we can solve this ODE using separation of variables.y&#039;+λy=0y&#039;=-λyIntegrating both sides with respect to x, we get:∫dyy=∫-λdxln|y|=-λx+CSolving for y, we get:y=Ce-λxUsing the initial condition&amp;nbsp;y(0)=0, we get:0=Ce-λ⋅00=CThus, the solution to the ODE with the given initial conditions is&amp;nbsp;y=0.

Let λ ∈ R be a parameter. We

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