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17-Peridot

## Nonlinear 1°ord differential equation

Hi everybody,
how can I solve this equation?

I found problems because of the derivative function squared. I need to collect the term derivative not squared...

Thanks
Bye

1 ACCEPTED SOLUTION

Accepted Solutions
16-Pearl
(To:gfraulini)

You can turn the non-linear y'(t)^2 + 4y'(t) = 5ty(t) into two simultaneous linear ode's:

Alan

9 REPLIES 9
23-Emerald III
(To:gfraulini)

Hmm squared.... Seems more like to the 3rd power.

You have (y'(t))^2 and mutiply that with 4 y'(t), that gives 4*(y'(t))^3...

Do you need a symbolic solution, or can you settle for a numeric approximation.

Either way: what are the initial values? [ y(0), y'(0)]

Luc

23-Emerald III
(To:LucMeekes)

Found it.

Try y(t)=+/-(1/8)*sqrt(10)*t^2.

The third solution is y(t)=0, but I guess you weren't looking for that.

Success!
Luc

24-Ruby V
(To:gfraulini)

Here's a possible workaround for a numerical solution, both Prime and real Mathcad:

17-Peridot
(To:gfraulini)

I'm so sorry because I wrote "(y'(t)^2) * 4y'(t)" instead of "(y'(t)^2) + 4y'(t)" that it changes a lot...

24-Ruby V
(To:gfraulini)

`The ODE must be linear in its highest derivative term,`

23-Emerald III
(To:gfraulini)

Does y(t) describe a physical phenomenon as a function of time? Or is this a made-up DE?

Luc

16-Pearl
(To:gfraulini)

You can turn the non-linear y'(t)^2 + 4y'(t) = 5ty(t) into two simultaneous linear ode's:

Alan

19-Tanzanite
(To:AlanStevens)

How to find y(0)=1?

Tokoro.

16-Pearl
(To:ttokoro)

y(0)=1 was an arbitrary choice on my part, as no initial conditions were specified in the original question.

z(0)=-4 gives rise to a rapidly decreasing y with time (and with y(0)=1). If this is what is expected from the model then fine! Personally, I think it’s probably safer to stick with z(0)=0.

Alan

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