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Given the system of equations $$\begin{cases} x'=x(x-1) \\ y'=-2xy+y \end{cases}$$ I am asked to calculate the general solution $\phi(t,0,x_0,y_0)$ that is, the general solution with initial conditions $\phi(0)=(x_0,y_0)$

Since I didn't know how to do it, I checked the solution and this was the first thing I saw:

$$\frac{x'}{x(x-1)}=1\implies \ln\bigg(\frac{|x-1|}{|x|}\bigg) - \ln\bigg(\frac{|x_0-1|}{|x_0|}\bigg)=t$$

Where does $$\ln\bigg(\frac{|x_0-1|}{|x_0|}\bigg)$$ come from?

I would have done $$\frac{x'}{x(x-1)}=1\implies \ln\bigg(\frac{|x-1|}{|x|}\bigg) - =t+c$$ and from here it's easy to find the explicit solution $x$.

I am studying dependence of solutions respect initial conditions and parameters of ordinary differential equations.

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  • $\begingroup$ Now just calculate from that last formula what $c$ is by inserting $(t,x)=(0,x_0)$. $\endgroup$ Dec 12, 2019 at 0:47
  • $\begingroup$ Well, let's solve a more general problem: $$ \begin{cases} x'\left(t\right)=x\left(t\right)\left(x\left(t\right)-\text{n}\right)\\ \\ y'\left(t\right)=y\left(t\right)-\text{k}x\left(t\right)y\left(t\right) \end{cases}\tag1 $$ Solving the first DE, we get: $$\frac{1}{\text{n}}\cdot\ln\left|\text{n}-\frac{\text{n}^2}{x\left(t\right)}\right|=t+\text{C}_1\tag2$$ Solving the second DE, we get: $$\ln\left|y\left(t\right)\right|=\text{C}_2+t-\text{k}\int x\left(t\right)\space\text{d}t\tag3$$ $\endgroup$ Dec 12, 2019 at 10:22

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$$\frac{x'}{x(x-1)}=1$$ can be written as $$\frac{\frac{dx}{dt}}{x(x-1)}=1$$ or $$\frac{dx}{x(x-1)}=dt$$ We can separate the left hand side:$$\frac{dx}{x-1}-\frac{dx}x=dt$$ When you integrate bot sides, you get $$\ln|x-1|-\ln|x|=t+c$$ Finally, using the properties of the logarithm $$\ln\left|\frac{x-1}x\right|=t+c$$ Plugging in $t=0$ and $x=x_0$ yields $$\ln\left|\frac{x_0-1}{x_0}\right|=c$$ Therefore $$\ln\left|\frac{x-1}x\right|=\ln\left|\frac{x_0-1}{x_0}\right|+t$$

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