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Mathematics 21 Online
OpenStudy (s):

use separation of variables to solve differential equation :dy/dx=t/(y+1)^(1/2) , y(1)=3 ?

sam (.sam.):

multiply both sides by\[\sqrt{y+1}\] Then integrate \[\frac{dy}{dx} \sqrt{y+1}\text{ = }t\] \[\sqrt{y+1}dy=tdx\] \[\int\limits_{}^{}\sqrt{y+1}dy=\int\limits_{}^{} tdx\]

OpenStudy (s):

I got this part, but I am confused how to solve it

sam (.sam.):

\[\int\limits \sqrt{y+1} \, dy=\frac{2}{3} (y+1)^{3/2}\] \[\int\limits_{}^{}tdx=tx+c\] ------------------------------------------------ \[\frac{2}{3} (y+1)^{3/2} =tx+c\] \[(y+1)^{3/2}=\frac{3(tx+c)}{2}\] \[y+1=(\frac{3(tx+c)}{2})^{2/3}\] \[y=(\frac{3(tx+c)}{2})^{2/3}-1\]

OpenStudy (anonymous):

I think it's supposed to be dy/dt ?

OpenStudy (s):

oops, yes, thats dy/dt

OpenStudy (anonymous):

then it's 2/3*(y+1)^(3/2) = (t^2)/2 +C

OpenStudy (anonymous):

use I.V. to find C , solve for y(t)

OpenStudy (s):

thanks !

OpenStudy (anonymous):

sure:)

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