Question

\[y' = x^3 (1-y)\]

Answer 1

this is easily separable...

Answer 2

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

Answer 3

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

Answer 4

ugh... check ur right hand side

Answer 5

Right hand side?
\[x^3 dx\]?

Answer 6

i think its fine
here u can isolate y easily.

Answer 7

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?

Answer 8

Hmm.. -ln2

Answer 9

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

Answer 10

but C would still be -ln2

Answer 11

It's x^3!

Answer 12

continue....

Answer 13

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}}}{

Answer 14

Sorry, please ignore the last line. OS just freezed my page.

Answer 15

But that doesn't look good at al..

Answer 16

better to do this
\( -\ln2+\ln |1-y| = -\frac{1}{4}(x)^4\)

Answer 17

\[-\ln2+\ln |1-y| = -\frac{1}{4}x^4\]\[\frac{1-y}{2}=e^{\frac{x^4}{4}}\]\[1-y=2e^{\frac{x^4}{4}}\]\[y=1-2e^{\frac{x^4}{4}}\]

Answer 18

e^{-...}

Answer 19

Oh!! \[e^{-\frac{x^4}{4}}\]

Answer 20

everything else is correct

Answer 21

Got it :)

Answer 22

Thanks!!

Answer 23

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

Answer 24

lol i like how they censor inappropriate words