0

so I'm given this differential equation

$$x'=x(1-x)$$

with the initial value being

$$x(0)=\frac{1}{10}$$

I have to find a solution for $x(t)$ of the differential equation

I can tell that the differential equation is separable

$$\frac{dx(t)}{dt} = x(t)(1-x(t))$$

$$\frac{dx(t)}{dt}=x(t)((-1-x(t))$$

then I procced to divide both side with the RHS

$$\frac{\frac{dx(t)}{dt}}{(x(t)-1)x(t)}=1$$

$$\int\frac{\frac{dx(t)}{dt}}{(x(t)-1)x(t)}dt=\int1 dt$$

I know that the RHS would give me $t+c_1$, after this, i'm pretty much stuck.

user8700908
  • 413

2 Answers2

2

HINT: use that $$\frac{1}{x(1-x)}=\frac{1}{x}-\frac{1}{x-1}$$

Dr. Sonnhard Graubner
  • 95,283
  • I should get $ln(|x|) - ln(|x-1|) + C_1 = t + C_2$ right? then take the exponential to solve for $x$ right?, but how do I get it for $x(t)$ – user8700908 Nov 04 '17 at 15:25
0

$$\frac {dx}{x (1-x)} = dt $$ $$\implies [\frac{1}{x} + \frac {1}{1-x}]dx = dt $$ $$\implies \frac {dx}{x} + \frac {dx}{1-x} = dt $$

Hope you can take it from here.