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HELP.. differential problem y' = (3x^2-1)/(3+2y)
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no, i need the value of y
difeq right? multiply by \[3+2y\] get \[(2y+3 )dy = x^3+1\] integrate both sides get \[y^2+3y=x^4+x+c\]
that was wrong
\[(2y+3 )dy = 3x^2+1\] \[y^2+3x=x^3+x+c\] is better, sorry about that
well except the plus x should be minus x i need more coffee, but i hope you get the idea. then solve for y
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thanks, this is really helpfull
solving is ok, right? use quadratic formula
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