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