# Homework Help: Analitical solution for dy/dt=ay^3+by^2+cy+d

1. Jul 25, 2012

### ced_the_jedi

Dear all

I have to solve an equation which seems to be simple:

It is dy/dt=ay^3+by^2+cy+d but my problem is that I am not able to give the analitical solution y(t)=... And I need your help for that...

What I have already done is :

- I know that you have to find the solutions of ay^3+by^2+cy+d=0 (the coefficients allow me to say that three solutions exist and lead to y1, y2 and y3 as real solutions)

- Then one can write :
dy/(ay^3+by^2+cy+d)=edy/(y-y1)+fdy/(y-y2)+gdy/(y-y3), with e, f, g easy to determine.

- Then you have to integrate on both sides of the equality : edy/(y-y1)+fdy/(y-y2)+gdy/(y-y3)=-dt

- which leads to something like that [(y-y1)^e][(y-y2)^f][(y-y3)^g]=exp(-t)

--> MY PROBLEM IS : do you know a way to express directly the analitical solution : y(t)=...? using this formulae, or by using a different integration method

2. Jul 25, 2012

### LCKurtz

I think you will get $e^{+t}$ on the right. To get an explicit solution instead of an implicit one, you need to solve a cubic equation for $y$. That can be done but except for special cases, it isn't pretty.

3. Jul 25, 2012

### jackmell

I think it's worst than that. He has:

$$(y-y_1)^a(y-y_2)^b(y-y_3)^c=e^t$$

No way you're getting a y out of there unless you're desperate and plot the data then invert the data and fit it to a nice-size polynomial.