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

Differential Equations
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this is easily separable...
\[y' = x^3(1-y)\]\[\frac{dy}{1-y} = x^3 dx\]\[-\ln |1-y| = \frac{1}{4}x^4+C\]So far so good?
lol I know, the whole chapter I'm doing now is separable.

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Other answers:

ugh... check ur right hand side
Right hand side? \[x^3 dx\]?
i think its fine here u can isolate y easily.
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?
Hmm.. -ln2
ok x^2 does not integrate to 1/4*x^4
but C would still be -ln2
It's x^3!
continue....
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}}}{
Sorry, please ignore the last line. OS just freezed my page.
But that doesn't look good at al..
better to do this \( -\ln2+\ln |1-y| = -\frac{1}{4}(x)^4\)
\[-\ln2+\ln |1-y| = -\frac{1}{4}x^4\]\[\frac{1-y}{2}=e^{\frac{x^4}{4}}\]\[1-y=2e^{\frac{x^4}{4}}\]\[y=1-2e^{\frac{x^4}{4}}\]
e^{-...}
Oh!! \[e^{-\frac{x^4}{4}}\]
everything else is correct
Got it :)
Thanks!!
\[\ln \left| 1-y \right| = -\frac{ 1 }{ 3 }x ^{3} + \ln2\] \[\left| 1-y \right| = 2e^{-\frac{ x ^{3} }{ 3 }}\] then you have to solve the absolute value part which gives u a piecewise function (this is a pain in the retrice
lol i like how they censor inappropriate words

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