## RolyPoly 3 years ago $y' = x^3 (1-y)$

1. cnknd

this is easily separable...

2. RolyPoly

$y' = x^3(1-y)$$\frac{dy}{1-y} = x^3 dx$$-\ln |1-y| = \frac{1}{4}x^4+C$So far so good?

3. RolyPoly

lol I know, the whole chapter I'm doing now is separable.

4. cnknd

ugh... check ur right hand side

5. RolyPoly

Right hand side? $x^3 dx$?

6. hartnn

i think its fine here u can isolate y easily.

7. RolyPoly

If it's fine, then I continue.. y(0) = 3 $-\ln |1-3| = \frac{1}{4}(0)^4+C$$-\ln |2| = C$ So far so good?

8. RolyPoly

Hmm.. -ln2

9. cnknd

ok x^2 does not integrate to 1/4*x^4

10. cnknd

but C would still be -ln2

11. RolyPoly

It's x^3!

12. hartnn

continue....

13. RolyPoly

So, $-\ln |1-y| = \frac{1}{4}(x)^4-\ln 2$$\ln2-\ln |1-y| = \frac{1}{4}(x)^4$$\ln\frac{2}{1-y} = \frac{1}{4}(x)^4$$\frac{2}{1-y} = e^{\frac{x^4}{4}}$$2= ({1-y} )e^{\frac{x^4}{4}}$$y = \frac{e^{\frac{x^4}{4}}-2}{\frace^{\frac{x^4}{4}}}{ 14. RolyPoly Sorry, please ignore the last line. OS just freezed my page. 15. RolyPoly But that doesn't look good at al.. 16. hartnn better to do this $$-\ln2+\ln |1-y| = -\frac{1}{4}(x)^4$$ 17. RolyPoly \[-\ln2+\ln |1-y| = -\frac{1}{4}x^4$$\frac{1-y}{2}=e^{\frac{x^4}{4}}$$1-y=2e^{\frac{x^4}{4}}$$y=1-2e^{\frac{x^4}{4}}$

18. hartnn

e^{-...}

19. RolyPoly

Oh!! $e^{-\frac{x^4}{4}}$

20. hartnn

everything else is correct

21. RolyPoly

Got it :)

22. RolyPoly

Thanks!!

23. cnknd

$\ln \left| 1-y \right| = -\frac{ 1 }{ 3 }x ^{3} + \ln2$ $\left| 1-y \right| = 2e^{-\frac{ x ^{3} }{ 3 }}$ then you have to solve the absolute value part which gives u a piecewise function (this is a pain in the retrice

24. cnknd

lol i like how they censor inappropriate words