The solution at x = 1, t = 1 of the partial differential equation, \(\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=25\frac{{{\partial }^{2}}u}{d{{t}^{2}}}\) subject to initial condition of \(u\left( 0 \right)=3x,\frac{\partial u}{\partial t}\left( 0 \right)=3\) is _____.

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  2. 2
  3. 4
  4. 6

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Option 4 : 6
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Concept:

It is 1D wave equation in the partial differential equation given by-

\(\frac{{{\partial }^{2}}\text{y}}{\partial {{\text{t}}^{2}}}={{\text{C}}^{2}}\frac{{{\partial }^{2}}\text{y}}{\partial {{\text{x}}^{2}}}\)

Where C2 = T/m, T = Tension in the elastic string, and M = mass per unit length

Calculation:

On comparing the above equation, we get that C = 1/5

f(x) = 3x, g(x) = 3

f (x + ct) = f (x + t/5) = 3 (x + t/5) = 3x + 3t/5

f (x - ct) = f (x - t/5) = 3 (x- t/5) = 3x - 3t/5

\(\text{U}(\left( \text{x},\text{t} \right)=\frac{1}{2}\left\{ \text{f}\left( \text{x}+\text{ct} \right)+\text{f}\left( \text{x}-\text{ct} \right)+\mathop{\int }_{\text{x}-\text{ct}}^{\text{x}+\text{ct}}\text{g}\left( \text{x} \right)\text{dx} \right\}\)

\(=\frac{1}{2}\left\{ 6x+\frac{1}{1/5}+\mathop{\int }_{x-t/5}^{x+t/5}3~dx \right\}\)

\(=\frac{1}{2}\left\{ 6\text{x}+5+\left[ \left( 3\text{x}+\frac{3\text{t}}{5} \right)-\left( 3\text{x}-\frac{3\text{t}}{5} \right) \right] \right\}\)

\(\text{U}\left( \text{x},\text{t} \right)=\frac{1}{2}\left( 6\text{x}+6\text{t} \right)\)

\(\therefore \text{U}\left( 1,1 \right)=\frac{1}{2}\left( 6\times 1+6\times 1 \right)=6\)

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