Solving ODE $x' = \lambda x^2$ I am currently studying continuous dependence ODE theory, and there's one example given in our lecture notes, where I am confused how to solve it.
The equation is: $\displaystyle x' = \lambda~x^2$ and furthermore there's a function defined as $\displaystyle f(t,x,\lambda) = \lambda~x^2$
Can anybody go with me through the steps to find the explicit solution in terms of $ \displaystyle t, t_0, x_0$ and $ \displaystyle \lambda$? I tried to solve the equation as a separable equation, but I wouldn't get to a satisfying result. 

Given $\dfrac{dx}{dt}=\lambda x^2,$ assuming $\lambda \in \mathbb{R}$, then 
$\dfrac{dx}{x^2}=\lambda dt$. Integrating both sides, we get $\int\dfrac{dx}{x^2}=\lambda \int dt$. 
This gives $-\dfrac{1}{x}=\lambda t +c$ as the answer. 
If its further given that $x(t_0)=x_0$ ($x_0\ne 0$), then applying the condition on the solution we get, 
$-\dfrac{1}{x_0}-\lambda t_0=c$. So the final solution becomes 
$-\dfrac{1}{x}=\lambda t -\dfrac{1}{x_0}-\lambda t_0$. Simplify this to get a better expression.
