I am studying how to solve certain nonlinear differential equations and am currently trying to solve the equation $\dot{x}^2-2x=1$, where $x$ is a function of $t$, with initial condition $x(1)=1$. My method is as follows:

Solving for $x$ in this equation gives $x=\frac{1}{2}(\dot x-1)$. Let $p=\dot x$, then differentiating with respect to $t$ gives $\frac{d}{dt}x=\dot x=\frac{d}{dt}\frac{1}{2}(\dot x-1)=\frac{1}{2}(2\ddot x\dot x)$ by chain rule, so we get the first-order differential equation $p=pp'$. If $p\equiv0$ then we get the solution $x(t)\equiv C$. Otherwise, dividing by $p$ gives $p'=1$, so $p=t+c_1$ and $x=\frac{t^2}{2}+c_1t+c_2$.

Plugging back into the original equation: $$(t+c_1)^2-2(\frac{t^2}{2}+c_1t+c_2)=1$$ $$t^2+2c_1t+c_1^2-t^2-2c_1t-2c_2=1$$ $$c_1^2-2c_2=1$$ $$c_1^2=1-2c_2$$ $$c_1=\pm\sqrt{1-2c_2}$$

Now applying the initial condition $x(1)=1$:


Solving this equation for $c_2$ I get $c_2=-\left(\sqrt5+\frac 5 2\right)$ and $c_2=\sqrt5-\frac 5 2$. Therefore, the solution is

$$x(t)=\frac{t^2}{2}\pm \sqrt{1+2(\frac 5 2\pm\sqrt5)}-\frac{5}{2}\pm\sqrt5$$

But according to WolframAlpha the solutions are $$x(t)=\frac{1}{2}\big(t^2\pm2(1+\sqrt3)t\mp(2\sqrt3)+3\big)$$

Where am I going wrong in my approach? Did I make an algebraic mistake or is there something more deeply wrong with this method?

  • $\begingroup$ If you meant what I think you did, and I would rather write as $\;x'-2x=1\;$ , with $\;x=x(t)\;$ a function of $\;t\;$, then this diff. eq. is linear . Did you mean perhaps something else? $\endgroup$
    – DonAntonio
    Apr 11 '19 at 18:53
  • $\begingroup$ @DonAntonio The $x'$ term is squared, so it is not linear. This is why I used the dot notation (to make it more readable than $(x')^2$). $\endgroup$
    – csch2
    Apr 11 '19 at 18:54
  • $\begingroup$ I completely missed that $\;2\;$ in the exponent. Thanks $\endgroup$
    – DonAntonio
    Apr 11 '19 at 18:57

It should be $c_1=\pm\sqrt{1+2c_2}$.

(BTW, I would solve this from $\dot{x}=\pm\sqrt{1+2x}$ and then separate the variables.)


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