Non-linear ode solution with integral formula I am trying to figure out how the ode $\dot{x}=x^2, x(0)=c>1$ has the solution $x(t)=(\frac{1}{c}-t)^{-1}$ using the formula $x(t) = x_0+ \int^{t}_{t_0}f(s,x(s))ds$. I tried the substitution suggested in the comments $u=\frac{1}{x}$ but it doesnt give the correct form. I've also tried thinking about linearization but cant figure out how it fits into the integral formula?
 A: Suppose we are given the initial value problem (IVP)
$$\dot{x}=f(t,x(t)),\quad x(0)=x_0\tag{1}$$
and the integral equation
$$x(t) = x_0+ \int^{t}_{0}f(s,x(s))\,ds\tag{2}$$
where $f$ is continuous in its domain. Then, $x(t)$ is a solution of the IVP $\iff$ $x(t)$ is a solution of the integral equation.
First, assume $x(t)$ is a solution to the integral equation. Then, 
$$\dot{x}=f(t,x(t)),\quad x(0)=x_0$$
It follows that
$$\frac{dx}{dt}=f(t,x(t))\implies\int_0^t dx=\int_0^t f(s,x(s))\,ds \implies x(t)-x(0)=\int_0^t f(s,x(s))\,ds $$
Next, assume that $x(t)$ is a solution to the integral equation
$$x(t) = x_0+ \int^{t}_{0}f(s,x(s))\,ds$$
Then, $f(s,x(s))$ is integrable from $0$ to $t$ for any $t$ in the domain. Thus, $\int_0^t f(s,x(s))\,ds$ is continuous in $t$. By $(2)$, we know that $x(t)$ is continuous. Moreover, since $f$ is continuous in its domain, $f(t,x(t))$ is continuous, from which it follows that the right-hand side of $(2)$ is differentiable, and so is $x(t)$. Differentiating both sides of $(2)$ we have $\dot{x}=f(t,x(t))$ where $x(0)=x_0$. Hence, $x(t)$ is a solution to the IVP $(1)$.
Given that $x(t)$ is a solution to both the IVP and the integral equation, we can work directly from $(1)$
$$\dot{x}=\frac{dx}{dt}=f(t,x(t))=x^2$$
which is a separable equation
$$\frac{1}{x^2}\,dx=dt$$
which forms
$$-x^{-1}+c_1=t \implies x(t)=\frac{1}{c_2-t}$$
the initial condition of $x(0)=c$ produces
$$c=\frac{1}{c_2}$$
therefore
$$x(t)=\frac{1}{\dfrac{1}{c}-t}=\left(\dfrac{1}{c}-t\right)^{-1}$$
