I have the following IVP:$$y^{\prime\prime}+a(y^{\prime})^2+by+c=0$$

with the following initial conditions:

\begin{cases} y(t=0)=h\\ y^{\prime}(t=0)=0\\ y^{\prime\prime}(t=0)=k \end{cases}

where, $a,b,c,h$ and $k$ are all non-null and known constants.

This is my actual problem:$$y''(t)=-100y(t)-0.01(y'(t))^2-10$$

With initial conditions:

\begin{cases} y(t=0) = -100\\ y'(t=0) = 0 \end{cases}

Where, $t$ represents time and starts at $t_0 = 0$.

I am interested in solving the problem from $t_0=0$ until $y(t)=0$.

Does this problem have a closed solution?

Thank you.

with WolframAlpha I obtained these plots, but I have no information on the solution:

enter image description here

  • $\begingroup$ Since this is a second-order DE, you have one too many initial conditions. $\endgroup$ – Robert Israel Sep 29 '13 at 4:26
  • $\begingroup$ @RobertIsrael the point is that this IVP comes from a physical problem which has indeed all those initial conditions... Should I re-formulate the IVP? $\endgroup$ – Alfredo Capobianchi Sep 29 '13 at 16:09
  • $\begingroup$ If you plug in $y(0) = h$ and $y'(0) = 0$ to the differential equation, you get $y''(0) = -b h - c$ $\endgroup$ – Robert Israel Sep 29 '13 at 18:59

By "analytical" I assume you mean closed-form. There is an implicit solution to the general differential equation involving an integral:

$$\int^{y(t)} \pm \frac{2a\ ds}{\sqrt{C a^2 \exp(-2sa) -4 s a b-4 ac+2b}} = t - t_0$$

In order for $y' = 0$ when $y = h$ you need $$C = {\frac { \left( 2\,hab+2\,ac-b \right) {{\rm e}^{2\,ha}}}{2{a}^{2} }}$$

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  • $\begingroup$ Thank you. The third initial condition can't be satisfied then? $\endgroup$ – Alfredo Capobianchi Sep 29 '13 at 16:10

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