## mays 3 years ago need help to solve this equation e^tdy/dt=y^3

1. sunnymony

e^tdy/dt=y^3 dyv^-3=dt e^-t on integratiion y^-2/2=e^-t/-1 +constant

2. mays

than you very much

3. sunnymony

ur welcome

4. mays

if i need the general solution of the differential equation expressing y explicitly as a function of t and hence find the particular solution that satisfies the initial y(0)=1 do i need to make it y=f(x) then find the constant c then substitute the value of x through f(x) thanks

5. mays

is it right

6. scarydoor

Have you written the question correctly? A first order non-linear differential equation? (non-linear?) Wolfram gives: http://www.wolframalpha.com/input/?i=e^t+dy%2Fdt%3Dy^3 A wacky solution... it seems to be interpreting it correctly, as least how I would think of it from what you've written...

7. mays

yes its true e^t *dy/dt=y^3 (where y is gretethan 0) but i use the wolframalpha but itesnt go right whatdo you thinease

8. mays

sorry what do you think please

9. scarydoor

http://www.wolframalpha.com/ Type into the search the problem: e^t *dy/dt=y^3 It will solve it. But it's not a nice looking solution...

10. mays

the answer i got is y=1/-2e^t+2c

11. mays

and this without computr

12. scarydoor

actually the solution the other guy gave is right. But it's written implicitly not explicity, which is why it looks different to what wolfram gives. But still the solution is very ugly. If this is for a class assignment then I think this is either written incorrectly, or it's supposed to look ugly for some reason (maybe it's from a real life problem).

13. mays

ok thanks lot will try it again

14. scarydoor

here's the solution that the other guy gave: -1/2 * y^(-2) = -e^(-t) + C. I think that's right. But see how that's written in implicit form. If you solve that for y then I think you should end up with what wolfram alpha got.