Why does $x_t = x^2$ imply $(\frac{1}{x(t,s)})_t=- 1 $?

Question

We have the following transport equation : $$x^2u_x-(y^2+1)u_y=ux^3$$

Use the method of characteristics to find a solution in $x > 1$ such that $u(1, y) = 2$.

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The characteristic equations and parametric initial conditions are given by

$$x_t(t,s)=x^2 \:\: y_t(t,s)=-(y^2+1) \:\: u_t(t,s)=ux^3$$ $$x(0,s)=1 \:\: y(0,s)=s \:\: u(0,s)=2$$

Solving for $x$ we get that $x_t = x^2$, so

$$(\frac{1}{x(t,s)})_t=- 1 $$ which implies $$\frac{1}{x(t,s)}- (\frac{1}{x(0,s)}) = -t$$

(...)

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I have two questions:
1) First, why does $x_t = x^2$ imply that $(\frac{1}{x(t,s)})_t=- 1 $ ?
2) And second, why do they integrate from $0$ to $t$ and not something else ?

Thanks for your help !

Answer 1

As already said in the comments by @fGDu94:

1) $x_t=x^2$ is a separable first order ODE, i.e. $1/x^2 dx= dt $. So $-1/x = t + (-C)$ i.e. $1/x = -t + C$ thus $1/(x(t,s)_t) = -1 + C$ where $C=1/(x(0,s))$ is a constant of integration.

2) Because we want to know the solution at time and we have an initial condition at time $0$, so we should integrate between these limits