Evaluation of $~{\mathrm{d}^2 x\over\mathrm{d}t^2}+\left({\mathrm{d}x\over\mathrm{d}t}\right)^2-4=0$

Question

$$ x=x(t):=C^{\infty}~\text{class function} $$

$$ \underbrace{\color{red}{{\mathrm{d}^2 x\over\mathrm{d}t^2}+\left({\mathrm{d}x\over\mathrm{d}t}\right)^2-4=0} }_{\text{I want to evaluate a general soln of this ODE}} $$

$$ x(0)={\mathrm{d}x\over\mathrm{d}t}(0)=0 $$

My tries:

NOTICED MISCALCULATION. I AM CORRECTING IT NOW

$$ \underbrace{{\mathrm{d}^2 x\over\mathrm{d}t^2}+\left({\mathrm{d}x\over\mathrm{d}t}\right)^2=4 }_{\text{Original ODE}} $$

$$ \underbrace{{\mathrm{d}^2 x\over\mathrm{d}t^2}+\left({\mathrm{d}x\over\mathrm{d}t}\right)^2=0 }_{\text{Prepared a new one}}\\ $$

I assume$~{\mathrm{d}x\over\mathrm{d}t}\neq0~$ for$~x\neq0$

$$\begin{align} {\mathrm{d}^2 x\over\mathrm{d}t^2}&=-\left({\mathrm{d}x\over\mathrm{d}t}\right)^2\\ \left({\mathrm{d}x\over\mathrm{d}t}\right)^{-1}{\mathrm{d}^2 x\over\mathrm{d}t^2}&=-\left({\mathrm{d}x\over\mathrm{d}t}\right)^1\\ \left({\mathrm{d}t\over\mathrm{d}x}\right){\mathrm{d}^2x\over\mathrm{d}t^2}&=-{\mathrm{d}x\over\mathrm{d}t}\\ \int\left(\left({\mathrm{d}t\over\mathrm{d}x}\right){\mathrm{d}^2x\over\mathrm{d}t^2}\right)\mathrm{d}t&=\int-{\mathrm{d}x\over\mathrm{d}t}~\mathrm{d}t\\ \int\left(\left({\mathrm{d}t\over\mathrm{d}x}\right){\mathrm{d}^2x\over\mathrm{d}t^2}\right)\mathrm{d}t&=-x+\text{const}\\ 　\left({\mathrm{d}x\over\mathrm{d}t}\right)\left({\mathrm{d}x\over\mathrm{d}t}\right)　-\int\left(\left({\mathrm{d}t\over\mathrm{d}x}\right){\mathrm{d}^2x\over\mathrm{d}t^2}\right)\mathrm{d}t&=-x+\text{const} \\ \left({\mathrm{d}x\over\mathrm{d}t}\right)^2　-\left\{\left({\mathrm{d}x\over\mathrm{d}t}\right)^2-\int\left(\left({\mathrm{d}t\over\mathrm{d}x}\right){\mathrm{d}^2x\over\mathrm{d}t^2}\right)\mathrm{d}t\right\}&=-x+\text{const}\\ \int\left(\left({\mathrm{d}t\over\mathrm{d}x}\right){\mathrm{d}^2x\over\mathrm{d}t^2}\right)\mathrm{d}t&=-x+\text{const} \end{align}$$

I've been stucked in the closed loop.

Where I've made (a) mistake(s)?

And how can I evaluate the general solution?

Answer 1

$${\mathrm{d}^2 x\over\mathrm{d}t^2}+\left({\mathrm{d}x\over\mathrm{d}t}\right)^2-4=0$$ Let $u={\dfrac{\mathrm{d}x}{\mathrm{d}t}}$, you have to solve: $$ u'+u^2= 4 \quad\quad (*)$$ You can solve $(*)$ using separation of variables.

Don't forget to go back to ${\dfrac{\mathrm{d}x}{\mathrm{d}t}}$ integrating $u$ ;).

Answer 2

By guessing and looking at the equation, try a solution of the form $x'=a\tanh bt$:

$$ab\operatorname{sech}^2bt+a^2\tanh^2bt -4 =0$$

Choose $a=b=2$ and you get

$$4\operatorname{sech}^2 2t + 4\tanh^2 2t - 4 = 4 - 4 =0$$

Since $\tanh(0)=0$, our solution is

$$x = \int 2\tanh 2t \:dt = \boxed{\log(\cosh 2t)}$$

which satisfies $x(0)=0$.

Answer 3

Hint The substitution $x = \log u$ transforms the initial value problem to a second-order linear i.v.p.

(One way to find this substitution is to observe that the original equation is a Riccati equation in the variable $y := x'$, namely $y' + y^2 - 4 = 0.$ The standard method for solving such equations applies the substitution $y = \frac{u'}{u} = (\log u)'$ to transform the equation into a second-order linear equation.)

The i.v.p. in $u$ is $$u'' - 4 u = 0, \qquad u(0) = 1, \qquad u'(0) = 0 .$$ The standard method for solving a second-order linear equation with constant coefficients gives the solution $$u = \cosh 2 t .$$