Question #8b242 Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Aug 29, 2016 # y =1 - 3/(4x^3 -1)# Explanation: #y' = 4x^2 (y-1)^2, y(0) =1# This is separable: #1/(y-1)^2 y' = 4x^2 # #int 1/(y-1)^2 y' dx =int 4x^2 dx# #int 1/(y-1)^2 dy =int 4x^2 dx# # -1/(y-1) = (4x^3)/3 + C# # -(y-1) = 1/((4x^3)/3 + C) = 3/(4x^3 + C)# # y =1 - 3/(4x^3 + C)# # y(1) = 0 = 1 - 3/(4 + C)# #4 + C = 3, C = -1# # y =1 - 3/(4x^3 -1)# 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 1315 views around the world You can reuse this answer Creative Commons License