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Mathematics 8 Online
OpenStudy (anonymous):

dy/dx= (5x^2)/sqrt(y) and y=1 when x=0 Use separation of variables to solve the initial value problem

zepdrix (zepdrix):

\[\Large\bf\sf \frac{dy}{dx}=\frac{5x^2}{\sqrt y},\qquad y(0)=1\]Ok what are you stuck on? :O

OpenStudy (anonymous):

ok when i do it i got y=((5x^3)/2)^3/2

OpenStudy (anonymous):

is that correct?

zepdrix (zepdrix):

Hmm let's check.

zepdrix (zepdrix):

\[\Large\bf\sf \int\limits \sqrt y \;dy\quad=\quad \int\limits 5x^2\;dx\]Gives us,\[\Large\bf\sf \frac{2}{3}y^{3/2}\quad=\quad \frac{5}{3}x^3+C\]

zepdrix (zepdrix):

Multiply both sides by 2/3,\[\Large\bf\sf y^{3/2}\quad=\quad \frac{5}{2}x^3+c\]

zepdrix (zepdrix):

Plug in our initial data,\[\Large\bf\sf 1\quad=\quad \frac{5}{2}\cdot 0+c\qquad\implies\qquad c=1\]

zepdrix (zepdrix):

Did you get your c value messed up maybe? Hmm

OpenStudy (anonymous):

i think that might be

OpenStudy (anonymous):

is it |dw:1392614228491:dw|

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