Solve differential equation with initial value $x(0)=-3$. I tried to solve it and came to the step which doesnt look good and I am not sure what to do next. Here are my steps:

$\frac{dx}{dt}=x^2+3x $

$\int_{}^{}\frac{dx}{x^2+3x} \! \, =\int_{}^{}dt \! \, $

$-\frac{1}{3}ln(1+\frac{3}{x} ) =t+C$

$x=\frac{3}{e^{-3(t+c)}-1 } $


$-3=\frac{3}{e^{-3c}-1} $



$$\frac{dx}{dt}=x^2+3x$$ $$\frac{dx}{x(x+3)}=dt$$ $$\frac{dx}{3x}-\frac{dx}{3(x+3)}=dt$$

  • $\begingroup$ Is the solution for constant: $3C=ln(1/6)$ ? $\endgroup$ – Ana Matijanovic Oct 14 '16 at 22:48

What you have so far looks correct.

What does it imply?

$c = \infty\\ e^{-3(c+t)}=0\\ x = -3$

You have an autonomous differential equation. That is, $x'$ depends solely on $x.$

Plug in the initial condition into the equation

$x'(0) = 0$

If $x'(0)$ then $x(\epsilon) = x(0) + \epsilon x'(0) = x(0)$

There is nothing to get this system moving. We never get off of the initial condition.

  • $\begingroup$ I dont understand what excatly here is $x'$ ? $\endgroup$ – Ana Matijanovic Oct 14 '16 at 22:49
  • $\begingroup$ $x' = \frac {dx}{dx}$ different notations for the same thing. $\endgroup$ – Doug M Oct 15 '16 at 0:15
  • $\begingroup$ But if plug x=-3 I am getting 0 in denominator? $\endgroup$ – Ana Matijanovic Oct 15 '16 at 8:05
  • $\begingroup$ Plug in x=-3 to the original $\frac {dx}{dt} = x^2 + 3x \implies \frac {dx}{dt} = 0$ No zeros in any denominators. $\endgroup$ – Doug M Oct 17 '16 at 15:32

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