How do I solve the following system of ODEs?


My idea was to use the 2nd equation to solve for $y_2$ (Riccati ODE). And then I would plug $y_2$ into the first equation and solve the first equation by separation.

  • 3
    $\begingroup$ It seems like you’re strategy should work. Did you have some problem implementing it? $\endgroup$
    – Robo300
    Jan 22, 2020 at 16:58
  • $\begingroup$ @Robo300 For the Riccati method I need a solution to start with. So I need to guess a solution for the 2nd ode. Can I simply choose $y_2=1$ because then we have $y_2'=1-y_2^2$? (both sides being equal to $0$). Or which solution should I take instead? $\endgroup$
    – user728962
    Jan 22, 2020 at 18:15
  • $\begingroup$ What makes this problem tractable is that, if you invert the second equation (trading $y_2$ for $t$) then what you have is just $t'(y_2) = (1-y_2)^{-2}$ which can be integrated directly. $\endgroup$ Jan 22, 2020 at 18:39

1 Answer 1


The second DE is separable you don't need any particular solution to solve it: $$\int \frac {dy_2}{1-y_2^2}=\int dt$$ You can use fraction decomposition: $$\int \frac {dy_2}{1-y_2^2}=\frac 1 2 \left (\int \frac {dy_2}{y_2+1}-\int \frac {dy_2}{y_2-1} \right )$$ Then solve first DE: $$\frac {y'_1}{y_1}=-y_2$$ $$(\ln y_1)'=-y_2$$

  • $\begingroup$ Okay so I get $\ln \big((\frac{y_2+1}{y_2-1})^{1/2}\big)=t+c \Rightarrow \frac{y_2+1}{y_2-1}=e^{2t}c \Rightarrow y_2=\frac{e^{2t}c+1}{e^{2t}c-1}$. Do I have to substitute this $y_2$ into the first equation or can I pick $c=1$? $\endgroup$
    – user728962
    Jan 22, 2020 at 19:44
  • $\begingroup$ Good job @user728962 yes you need to substitute that function in the first equation and get $y_1$ $\endgroup$ Jan 22, 2020 at 19:50
  • $\begingroup$ You need to have two constants so keep the constant as it is @user728962 $\endgroup$ Jan 22, 2020 at 20:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .