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Chapter 10: Problem 8
\(f(x)=\pi-x, \quad 0
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$$\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}, 0<\mathcal{x}<\pi,t>0,$$ $$u(0, t)=0, \quad u(\pi, t)=3 \pi, \quad t>0,$$ $$u(x, 0)=0,0<\mathcal{x}<\pi$$
\(f(x, y)=\cos 6 x \sin 4 y-3 \cos x \sin 11 y\)
$$f ( x ) = \sin x - 7 \sin 3 x + \sin 5 x$$
Verify d' Alembert's solution \((32)\) to the initial value problem \((26)-(28)\) when \(f(x)\) has a continuous second derivative and \(g(x)\) has a continuous first derivative by substituting it directly into the equations. $$ \frac{\partial^{2} u}{\partial t^{2}} $$ \(=\alpha^{2} \frac{\partial^{2} u}{\partial x^{2}}\), \(-\infty<\)x\(<\infty\), \(t>0\), $$ u(x, 0)=f(x) $$ \(-\infty<\)x\(<\infty\), $$ \frac{\partial u}{\partial t}(x, 0) $$, \(=g(x)\), \(-\infty<\)x\(<\infty\) for the given functions \(f(x)\) and \(g(x)\)
\(\frac{\partial^{2} u}{\partial t^{2}}=\frac{\partial^{2} u}{\partial x^{2}}\) 0\(<\)x\(<\)1 t\(>\)0 $$ u(0, t)=u(1, t)=0, \quad t>0$$ $$ u(x, 0) $$$=x(1-x)$$0\(<\)x\(<\)1\( $$ \frac{\partial u}{\partial t}(x, 0) $$$=\sin 7 \pi x$$0\)<\(x\)<\(1\)
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