the ends of a string of length L are secured at both ends. Its initial shape is straight but it is vibrating with initial velocity f(x)=x(L-x).choose the PDE and boundary/initial conditions that model this scenario. Find Eigenfunctions for ).The problem splits into cases based on the sign of A(Notation: For the cases below, use constants a and b)• Case 1: A = 0X(x)Plugging the boundary values into this formula gives0 = X(0)%3D0 = X(6)So X(x) =which means u(x, t) =We can ignore this case.• Case 2: A =-y² <0(In your answers below use gamma instead of lambda)X(x) =Plugging the boundary values into this formula gives0 – X(0)0 = X(6) =So X(x) =which means u(x, t) =We can ingore this case.• Case 3: A= >X(x) =(In your answers below use gamma instead of lambda)Plugging in the boundary values into this formula gives0 = X(0) =%3D0 = X(6)Which leads us to the eigenvalues A, = Yn where Yn =and eigenfunctions X,(z) =(Notation: Eigenfunctions should not include any constants a or b.)

Let the given equation be, x''-λx=0.

The problem splits into cases based on the sign of ). (Notation: For the cases below, use constants a and b) • Case 1: A = 0 X(z) = %3D Plugging the boundary values into this formula gives 0 X(0) = %3D %3D 0 = X(6) = So X(x) = %3D which means u(r, t) = %3D We can ignore this case. Case 2: A = -y? < 0 (In your answers below use gamma instead of lambda) X(x) = %3D Plugging the boundary values into this formula gives 0 = X(0) = 0 = X(6) So X(x) = %3D %3D which means u(x, t) = We can ingore this case. Case 3: A = y >0 (In your answers below use gamma instead of lambda) X(x) = %3D Plugging in the boundary values into this formula gives 0 = X(0) = 0 = X(6) =

The problem splits into cases based on the sign of ). (Notation: For the cases below, use constants a and b) • Case 1: A = 0 X(z) = %3D Plugging the boundary values into this formula gives 0 X(0) = %3D %3D 0 = X(6) = So X(x) = %3D which means u(r, t) = %3D We can ignore this case. Case 2: A = -y? < 0 (In your answers below use gamma instead of lambda) X(x) = %3D Plugging the boundary values into this formula gives 0 = X(0) = 0 = X(6) So X(x) = %3D %3D which means u(x, t) = We can ingore this case. Case 3: A = y >0 (In your answers below use gamma instead of lambda) X(x) = %3D Plugging in the boundary values into this formula gives 0 = X(0) = 0 = X(6) =

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

Question

the ends of a string of length L are secured at both ends. Its initial shape is straight but it is vibrating with initial velocity f(x)=x(L-x).

choose the PDE and boundary/initial conditions that model this scenario.

Find Eigenfunctions for ).
The problem splits into cases based on the sign of A
(Notation: For the cases below, use constants a and b)
• Case 1: A = 0
X(x)
Plugging the boundary values into this formula gives
0 = X(0)
%3D
0 = X(6)
So X(x) =
which means u(x, t) =
We can ignore this case.
• Case 2: A =-y² <0
(In your answers below use gamma instead of lambda)
X(x) =
Plugging the boundary values into this formula gives
0 – X(0)
0 = X(6) =
So X(x) =
which means u(x, t) =
We can ingore this case.
• Case 3: A= >
X(x) =
(In your answers below use gamma instead of lambda)
Plugging in the boundary values into this formula gives
0 = X(0) =
%3D
0 = X(6)
Which leads us to the eigenvalues A, = Yn where Yn =
and eigenfunctions X,(z) =
(Notation: Eigenfunctions should not include any constants a or b.)

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