For the general solution of the differential equation in X use A and B for your constants and list the functions in alphabetical order, for example y=????cos(x)+????sin(x)y=Acos⁡(x)+Bsin⁡(x). For the differential equation in T use the C and D.For the variable ????λ type the word lambda and type alpha for ????α,otherwise treat them as you would any other variable.
Use the prime notation for derivatives, so the derivative of ????X is written as ????′X′. Do NOT use ????′(x)X′(x)
The longitudinal displacement u(x,t) of a vibrating elastic bar can be modeled by a wave equation with free-end conditions
????2∂2????∂x2=∂2????∂????2,00a2∂2u∂x2=∂2u∂t2,00
∂????∂x∣∣∣x=0=0,∂????∂x∣∣∣x=????=0,????>0∂u∂x|x=0=0,∂u∂x|x=L=0,t>0
????(x,0)=x∂????∂????∣∣∣????=0=−2,0

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abhinavjh321 abhinavjh321 · Answer 1 · 2022-12-19T05:25:10+01:00

This differential equation's general answer is

u(x,t)=Acos⁡(αx−ωt)+Bsin⁡(αx−ωt)+Cx+D

The wave equation with free-end conditions is a differential equation of the form

????2∂2????∂x2=∂2????∂????2,00a2∂2u∂x2=∂2u∂t2,00

where ???? is the longitudinal displacement of a vibrating elastic bar, and x and t are the spatial and temporal variables, respectively.

∂????∂x∣∣∣x=0=0,∂????∂x∣∣∣x=????=0,????>0∂u∂x|x=0=0,∂u∂x|x=L=0,t>0

and the initial condition is

????(x,0)=x∂????∂????∣∣∣????=0=−2,0

u(x,0)=x,∂u∂t|t=0=-2,0

To solve this differential equation, we use the method of separation of variables. We first rewrite the equation as

a2∂2u∂x2=∂2u∂t2,

and then we assume that u can be written as a product of two functions, one of x and one of t. That is,

u(x,t)=X(x)T(t).

Substituting this into the wave equation and rearranging, we obtain two equations for X and T:

a2X″(x)=−ω2T(t)

a2T″(t)=−ω2X(x).

X(x)=Acos⁡(αx)+Bsin⁡(αx).

T(t)=Ccos⁡(ωt)+Dsin⁡(ωt).

Therefore, the general solution of the wave equation with free-end conditions is

u(x,t)=Acos⁡(αx−ωt)+Bsin⁡(αx−ωt)+Cx+D.

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