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cnknd Group TitleBest ResponseYou've already chosen the best response.1
this is easily separable...
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
\[y' = x^3(1y)\]\[\frac{dy}{1y} = x^3 dx\]\[\ln 1y = \frac{1}{4}x^4+C\]So far so good?
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
lol I know, the whole chapter I'm doing now is separable.
 2 years ago

cnknd Group TitleBest ResponseYou've already chosen the best response.1
ugh... check ur right hand side
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Right hand side? \[x^3 dx\]?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
i think its fine here u can isolate y easily.
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
If it's fine, then I continue.. y(0) = 3 \[\ln 13 = \frac{1}{4}(0)^4+C\]\[\ln 2 = C\] So far so good?
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Hmm.. ln2
 2 years ago

cnknd Group TitleBest ResponseYou've already chosen the best response.1
ok x^2 does not integrate to 1/4*x^4
 2 years ago

cnknd Group TitleBest ResponseYou've already chosen the best response.1
but C would still be ln2
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
continue....
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
So, \[\ln 1y = \frac{1}{4}(x)^4\ln 2\]\[\ln2\ln 1y = \frac{1}{4}(x)^4\]\[\ln\frac{2}{1y} = \frac{1}{4}(x)^4\]\[\frac{2}{1y} = e^{\frac{x^4}{4}}\]\[2= ({1y} )e^{\frac{x^4}{4}}\]\[y = \frac{e^{\frac{x^4}{4}}2}{\frace^{\frac{x^4}{4}}}{
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Sorry, please ignore the last line. OS just freezed my page.
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
But that doesn't look good at al..
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
better to do this \( \ln2+\ln 1y = \frac{1}{4}(x)^4\)
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
\[\ln2+\ln 1y = \frac{1}{4}x^4\]\[\frac{1y}{2}=e^{\frac{x^4}{4}}\]\[1y=2e^{\frac{x^4}{4}}\]\[y=12e^{\frac{x^4}{4}}\]
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Oh!! \[e^{\frac{x^4}{4}}\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
everything else is correct
 2 years ago

cnknd Group TitleBest ResponseYou've already chosen the best response.1
\[\ln \left 1y \right = \frac{ 1 }{ 3 }x ^{3} + \ln2\] \[\left 1y \right = 2e^{\frac{ x ^{3} }{ 3 }}\] then you have to solve the absolute value part which gives u a piecewise function (this is a pain in the retrice
 2 years ago

cnknd Group TitleBest ResponseYou've already chosen the best response.1
lol i like how they censor inappropriate words
 2 years ago
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