How do you find the particular solution to #sqrtx+sqrtyy'=0# that satisfies y(1)=4? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jan 24, 2017 #implies y = (9 - x^(3/2)) ^(2/3)# Explanation: #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)# Answer link Related questions How do you solve separable differential equations? How do you solve separable first-order differential equations? How do you solve separable differential equations with initial conditions? What are separable differential equations? How do you solve the differential equation #dy/dx=6y^2x#, where #y(1)=1/25# ? How do you solve the differential equation #y'=e^(-y)(2x-4)#, where #y5)=0# ? How do you solve the differential equation #(dy)/dx=e^(y-x)sec(y)(1+x^2)#, where #y(0)=0# ? How do I solve the equation #dy/dt = 2y - 10#? Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the... How do I solve the differential equation #xy'-y=3xy, y_1=0#? See all questions in Solving Separable Differential Equations Impact of this question 2867 views around the world You can reuse this answer Creative Commons License