How do you find the particular solution to #sqrtx+sqrtyy'=0# that satisfies y(1)=4?

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Answer 1 · 2017-01-24T22:09:00

#implies y = (9 - x^(3/2)) ^(2/3)#

#sqrtx+sqrty \ y'=0#

This is separable:

#sqrty \ y'= - sqrtx#

#\int \ dy \ sqrty= - int \ dx \ sqrtx #

#2/3 y^(3/2) = - (2/3 x^(3/2) + C) #

#implies y^(3/2) = - x^(3/2) + C #

#y(1) = 4 implies (y(1))^(3/2) = 8 =-1 + C implies C = 9#

#implies y = (9 - x^(3/2)) ^(2/3)#