Acceleration is given in terms of position then it has been asked to find the position at time t

Question

The question is as follows :

The acceleration $a$ of a particle is given as $a(x) = x$ where $x$ is the position of the object on the $x$-axis. Then find the position of the object in terms of time $t$ i.e., find $x(t)$. It has been given that $x(t=0)=0$ i.e., the object is at the origin at time $t = 0$.

My attempt so far:

We can relate velocity and acceleration as functions of position using $v = \frac{dx}{dt}$ and $a = \frac{dv}{dt}$

Dividing, we get $v\ dv = a \ dx$. Now we can integrate $v$ and $a$ from $x = 0$ to $x$ by putting $a = x$ in the above differential equation.

Therefore, $v(x) = x$. Now if we would use $v = \frac{dx}{dt}$ and integrate $x$ from $0$ to $x$, then

$$x = \frac{dx}{dt}\\ \frac{dx}{x} = dt\\ \ln|x|\bigg{|}_0^x=t $$

here, $\ln(x)$ is undefined at the lower limit $x = 0$.

So, how can we move forward from this?

Answer 1

You need an initial velocity. Otherwise, if the object starts at $x=0$ then $a=x=0$. Since $v=0$, there will be no motion.

Answer 2

An alternative approach:

You are being asked to find a family of solutions to the linear differential equation

$$\displaystyle \frac {d^2x}{dt^2} = x(t)$$

with the condition that $x(0)=0$.

If

$$\displaystyle v(t) = \frac {dx}{dt} = kx(t)$$

then

$$\displaystyle \frac {d^2x}{dt^2} = k \frac {dx}{dt} = k^2x(t)$$

So to make $k^2=1$ we must have $k=\pm 1$. We therefore have a two parameter family of solutions:

$$x(t) = Ae^t + Be^{-t}$$

The constants $A$ and $B$ are constrained by the condition that $x(0)=0$, but you will be left with one free parameter, which you can relate to the value of $v(0)$.