Given the standard wave equation for small amplitudes, we have been asked to find the position of string $y(x,t)$, given:
$y(x,0)=\sin x$, and,
$y'(x,0)=\cos x$, where $y'$ depicts partial derivative of $y$ wrt $t$.
Please, someone help me realise the situation, and supply a solution to this problem, as well.
The argument of a moving wave will be of the form $\omega(t- \frac xv)$, where $ \omega$ is the time frequency and $v$ is velocity.
$$Y(x,t)=X(x)T(t)$$
Insert into the wave equation:
$$X''T=\frac{XT''}{v^2}$$
Where $X''$ and $T''$ are derivatives wrt to their own variable. Rearrange:
$$\frac{X''}{X}=\frac{T''}{v^2T}=-\omega^2$$
where $-\omega^2$ is the constant of separation. Since both sides are functions of only one variable, we can treat both sides as ordinary differential equations.
$$X=\{\sin(\omega \ x),\cos(\omega \ x) \}$$
$$T=\{\sin(\omega\ v\ t),\cos(\omega\ v\ t) \}$$
$Y$ will be a linear combination of these linearly independent solutions
$$Y=A\sin(\omega \ x)\sin(\omega\ v\ t)+B\sin(\omega \ x)\cos(\omega\ v\ t)\\ +C\cos(\omega \ x)\sin(\omega\ v\ t)+D\cos(\omega \ x)\cos(\omega\ v\ t) $$
It is clear that we have $\omega=1$
At $t=0$ we have $Y=\sin x$ so $D=0$ and $B=1$
At $t=0$ we have
$$\frac{\partial}{\partial t}Y=vA\sin x \cos vt + \sin x \sin vt+ vC\cos x \cos vt=\cos x$$
so we get $C=\frac1v$ and $A=0$
the result is
$$Y=\sin x \cos vt + \frac1v \cos x \sin vt$$