What is the answer to $\int x(t)dt$?

Question

$\int x(t)dt$?

I'm trying to solve a differential equation, but I've hit a strange brick wall that I never used to have a problem climbing over.

This question is about mechanics & the equation of motion.

I'm trying to solve $\frac{dx}{dt}=-bx$ where $b>0$.

So by integrating both sides, I get:

$x=-\frac{1}{2}bx^2+C$, where C is the constant of integration.

Am I doing this correctly, or is it supposed to be $x=-bxt$?

I apologise for how simple this question is, but I can't get my head around it.

Answer 1

Hint

Your differential equation is separable. Just rewrite $$\frac{dx}{dt}=-bx$$ as $$\frac{dx}{x}=-b \text { } dt$$ and integrate both sides (do not forget the integration constant).

I am sure that you can take from here.

Answer 2

Your first option is wrong because at left side you integrate respect to $t$ and at right side respect to $x$. And second option is wrong too, because $x$ depends on $t$.

A correct way is: $$x'(t)=-bx(t)$$ so $$\frac{x'(t)}{x(t)}=-b$$ Now integrate respect to $t$ to get $$\ln x(t)=-bt+C$$