Question

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

• A. $1$
• B. $2$
• C. $4$
• D. $6$

Answer 1

The correct option is D $6$

Standard form of wave equation is
$\frac{{\partial }^{2}u}{\partial t^2}=C^2\frac{{\partial }^{2}u}{\partial x^2}$
Given wave equation is
$\frac{{\partial }^{2}u}{\partial x^2}=25\frac{{\partial }^{2}u}{\partial t^2}$
$\Rightarrow \frac{{\partial }^{2}u}{\partial t^2}=\frac{1}{25}\frac{{\partial }^{2}u}{\partial x^2}$ so, $C=\frac{1}{5}$ and $f\left(x\right)=3x$ and $g\left(y\right)=3$
So by De-alemberts solution of wave equation..
$u(x,t)=\frac{1}{2}\left[f\right(x+ct)+f(x-ct\left)\right]+\frac{1}{2c}{\int }_{x-ct}^{x+ct}g\left(y\right)dy]$
$=\frac{1}{2}[3(x+\frac{1}{5}t)+3(x-\frac{1}{5}t)]+\frac{5}{2}{\int }_{x-ct}^{x+ct}3dy$
$=\frac{1}{2}\left[6x\right]+\frac{15}{2}[x+ct-x+ct]=3x+\frac{15}{2}\left(\frac{2}{5}t\right)$
$u(x,t)=3x+2t$
So, $u(1,1)=3+3=6$