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Differential Equation y'=3y^(2/3) and y(2)=0
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\[y'=3\sqrt[3]{y^2}\]
\[y(2)=0\]
the final answer is \[y=(x-2)^3\]
if you guys could help me out with it it will literally
make my day*
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Have you considered the antiderivative? \(y = 3 \dfrac{y^{5/3}}{5/3} + C\) Simplify first, of course.
wow
is it the same? can you explain?
@shereenkhan ?
nah @tkhunny I don't think that's the way tbh
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what do you have to do?
to get y need to integrate y'
are you sure?
wby not just writing instead of y'=dy/dx?
|dw:1427457678079:dw|
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