Laplace transform of the square of first derivative I want to compute the laplace transformation of this equation:
$$4 -0.1{\operatorname{d}\!x(t)\over\operatorname{d}\!t} - 0.01\left({\operatorname{d}\!x(t)\over\operatorname{d}\!t}\right)^2 = 0.9{\operatorname{d^2}\!x(t)\over\operatorname{d}\!t^2}$$ 
But I don't know what I have to do with the term $\displaystyle \left({\operatorname{d}\!x(t)\over\operatorname{d}\!t}\right)^2$. Are there rules to compute the second power of a derivative?
 A: The ODE is not a linear one. The Laplace transform method is not convenient. It is simpler to change of function first :
$${\operatorname{d}\!x(t)\over\operatorname{d}\!t}=Y(t)$$
$$4 -0.1\: Y(t) - 0.01 \big(Y(t)\big)^2 = 0.9{\operatorname{d}\!Y(t)\over\operatorname{d}\!t}$$
Then, continue with the usual method to solve the ODE of separable kind.
METHOT TO SOLVE YOUR PROBLEM :
1- Solve the above equation for $Y(t)$. So, you obtain explicitly $\frac{dx(t)}{dt}$.
2- Put $\quad Y(t)\quad $ into your second equation $\quad 0.9\frac{d^2y}{dt^2}=0.13\left(Y(t)\right)^2-9\quad $. 
So, you obtain $\quad \frac{d^2y}{dt^2}=$ an explicit fonction of $t$.
3- Integrate two times this function of $t$ which leads to $y(t)$.
NOTE : Since the initial system of two equations includes numerical values which look like coming from a physical problem, the analytical solving might be not the simplest method. The above result on the form of functions is complicated for further use in numerical applications. Why not solving the ODEs directly with numerical calculus ?
