Question #162010

(D^2 - α^2D'^2)z =x


1
Expert's answer
2021-02-24T12:25:52-0500

A partial solution of an equation utt(x,t)a2uxx(x,t)=f(x,t)u_{tt}(x,t) – a^2u_{xx}(x,t)=f(x,t) with the initial conditions u(x,t)=0u(x,t)=0 is given as follows:

u(x,t)=12a0txa(tτ)x+a(tτ)f(ξ,τ)dξdτu(x,t)=\frac{1}{2a}\int\limits_{0}^{t}\int\limits_{x-a(t-\tau)}^{x+a(t-\tau)}f(\xi,\tau) d\xi d\tau

Replacing f(x,t)f(x,t) with xx, we have

u(x,t)=12a0txa(tτ)x+a(tτ)ξdξdτ=12a0t((x+a(tτ))22(xa(tτ))22)dτ=0tx(tτ)dτ=xt2/2u(x,t)=\frac{1}{2a}\int\limits_{0}^{t}\int\limits_{x-a(t-\tau)}^{x+a(t-\tau)}\xi d\xi d\tau = \frac{1}{2a}\int\limits_{0}^{t} (\frac{(x+a(t-\tau))^2}{2}-\frac{(x-a(t-\tau))^2}{2}) d\tau = \int\limits_{0}^{t} x(t-\tau)d\tau=xt^2/2 The general solution of a linear non-homogeneous equation is a sum of a partial solution and a general solution of the homogeneous equation.

The final answer will be

u(x,t)=xt2/2+g(xat)+h(x+at)u(x,t)=xt^2/2 + g(x-at)+h(x+at)


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