Find the general solution of : $$\frac{dx}{dt}=-2x$$
i) By inspection trial solution
ii) By separating variables
iii) Verify that your answer is correct
I tried to integrate it by multiply both side by dt and I acquire $x(t)=-2xt$ but the answer provided is $x(t)=e^{-2t}$
And for part 2 the provided answer given is $x=e^{-2t}+c$ but i got is
It seems like you are treating $x$ as a constant with respect to $t$, but $x$ is a function of $t$. When you solve by separation of variables you get $$\frac{dx}{dt}=-2x \Rightarrow \frac{dx}{x}=-2dt $$ and integrating you find $$\log \lvert x \rvert=\int \frac{dx}{x} = -2\int dt=-2t+c.$$ Hence $\lvert x \rvert = e^{-2t+c}$.