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Question Number 100891 by bemath last updated on 29/Jun/20

$${\mathrm{u}}_{\mathrm{tt}}={\mathrm{u}}_{\mathrm{xx}}-6\mathrm{x};0⩽\mathrm{x}<\pi ,\mathrm{t}>0$$ $${\mathrm{u}}_{(0,\mathrm{t})}=0;{\mathrm{u}}_{(\pi ,\mathrm{t})}={\pi }^3+3\pi$$ $${\mathrm{u}}_{(\mathrm{x},0)}={\mathrm{x}}^3+3\mathrm{x}+3\sin \mathrm{x}$$ $${\mathrm{u}}_{\mathrm{t}}(\mathrm{x},0)=0$$

Answered by bramlex last updated on 29/Jun/20

$$u_{xx}-u_{tt}=6x\Rightarrow (D_x^2-D_t^2)u=6x$$ $$u=\frac{1}{D_x^2}(1-\frac{D_t^2}{D_x^2})\left(6x\right)$$ $$\to Taylorseries{(1-x)}^{-1}=1+x+x^2+x^3+...$$ $$u=\frac{1}{D_x^2}(1+\frac{D_t^2}{D_x^2}+...)\left(6x\right)$$ $$u=\frac{1}{D_x^2}(6x+0+...)$$ $$u=x^3+g_1\left(t\right)x+g_2\left(t\right)$$ $$u(0,t)\mathrm{\&}u(\pi ,t)\to g_1\left(t\right)=3,g_2\left(t\right)=0$$ $$u_1(x,t)=x^3+3x$$ $$u_{xx}-u_{tt}=0,setu=\mathrm{\varnothing }(x+at)$$ $${\mathrm{\varnothing }}^{″}(x+at)\times a^2=0\to 1-a^2=0$$ $$a=\pm 1\to u_2(x,t)=\mathrm{\varnothing }(x+t)+\mathrm{\varnothing }(x-t)$$ $$u=u_1+u_2\because u(x,t)=\mathrm{\varnothing }(x+t)+\mathrm{\varnothing }(x-t)+x^3+3x$$ $$sou(x,0)=x^3+3x+3\sin x$$ $$2\mathrm{\varnothing }\left(x\right)=3\sin x\to \mathrm{\varnothing }(x\pm t)=\frac{3}{2}\sin (x\pm t)$$ $$u(x,t)=\frac{3}{2}\sin (x+t)+\frac{3}{2}\sin (x-t)+x^3+3x★$$ $$$$

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