What is a solution to the differential equation #sqrtx+sqrtyy'=0# with y=4 when x=1?

Question

1464 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-07T21:58:43

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

#sqrtx+sqrty \ y'=0#

this can be separated

#sqrty \ y'=- sqrtx#

#int sqrty \ y' \ dx =-int sqrtx \ dx#

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

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

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

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

applying the IV:#y(1) = 4#

# 4 =(C - 1^(3/2))^(2/3) #

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

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