- #1

- 2,440

- 316

- TL;DR Summary
- In determining the solutions; would i be correct to to say that one has to first solve the differential equation to get the function then only then can one determine the solutions by checking on the family of curves- substituting for ##c## values.

I am going through this,

I noted that, i shall have a separation of variables, that leads to

$$\left[\int \dfrac{1}{y(y-1)} dy\right]= \int \dfrac {1}{6} dt$$

and using partial fraction, i then have,

$$\left[\int -\dfrac{1}{y} dy - \int \dfrac{1}{y-1} dy\right] = \int \dfrac {1}{6} dt$$

$$-(\ln y + \ln (y-1) = \dfrac{1}{6} t + c$$

$$-\ln y(y-1) = \dfrac{1}{6} t + c$$

$$\ln y(y-1) = -\left[\dfrac{1}{6} t + c\right]$$

$$e^{{-\frac{1}{6} t - c}} = y^2-y$$

which is an implicit solution to get an explicit solution one may solve the quadratic equation,

$$y^2-y-e^{-\frac{1}{6} t - c}=0$$

...

Edit;

I just amended ##x## to ##t## so as to have an autonomous ode.

... from this point ( that is, at the implicit solution) i was able to get the shape of the function. Using different assigned values of ##c## should give me the directional field that is shown... unless there is another approach then i stand to be informed hence reason of this post.

My specific question is this,

Do we have other ways of determining the nature of solutions other than first determining the integral curves?

The other aspect of checking for stability of soluitions ( asymptotically or unstable) is pretty straightforward. Cheers.

I noted that, i shall have a separation of variables, that leads to

$$\left[\int \dfrac{1}{y(y-1)} dy\right]= \int \dfrac {1}{6} dt$$

and using partial fraction, i then have,

$$\left[\int -\dfrac{1}{y} dy - \int \dfrac{1}{y-1} dy\right] = \int \dfrac {1}{6} dt$$

$$-(\ln y + \ln (y-1) = \dfrac{1}{6} t + c$$

$$-\ln y(y-1) = \dfrac{1}{6} t + c$$

$$\ln y(y-1) = -\left[\dfrac{1}{6} t + c\right]$$

$$e^{{-\frac{1}{6} t - c}} = y^2-y$$

which is an implicit solution to get an explicit solution one may solve the quadratic equation,

$$y^2-y-e^{-\frac{1}{6} t - c}=0$$

...

Edit;

I just amended ##x## to ##t## so as to have an autonomous ode.

... from this point ( that is, at the implicit solution) i was able to get the shape of the function. Using different assigned values of ##c## should give me the directional field that is shown... unless there is another approach then i stand to be informed hence reason of this post.

My specific question is this,

Do we have other ways of determining the nature of solutions other than first determining the integral curves?

The other aspect of checking for stability of soluitions ( asymptotically or unstable) is pretty straightforward. Cheers.

Last edited: