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dy/dx= (5x^2)/sqrt(y) and y=1 when x=0 Use separation of variables to solve the initial value problem
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\[\Large\bf\sf \frac{dy}{dx}=\frac{5x^2}{\sqrt y},\qquad y(0)=1\]Ok what are you stuck on? :O
ok when i do it i got y=((5x^3)/2)^3/2
is that correct?
Hmm let's check.
\[\Large\bf\sf \int\limits \sqrt y \;dy\quad=\quad \int\limits 5x^2\;dx\]Gives us,\[\Large\bf\sf \frac{2}{3}y^{3/2}\quad=\quad \frac{5}{3}x^3+C\]
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Multiply both sides by 2/3,\[\Large\bf\sf y^{3/2}\quad=\quad \frac{5}{2}x^3+c\]
Plug in our initial data,\[\Large\bf\sf 1\quad=\quad \frac{5}{2}\cdot 0+c\qquad\implies\qquad c=1\]
Did you get your c value messed up maybe? Hmm
i think that might be
is it |dw:1392614228491:dw|
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