What is a solution to the differential equation #y'=-9x^2y^2#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jul 31, 2016 # y =1/ (3 x^3 +C)# Explanation: #y'=-9x^2y^2# this is separable #1/y^2 y'=-9x^2# #int \ 1/y^2 y' \ dx=int \ -9x^2 \ dx# #int \ 1/y^2 \ dy=-9 int \ x^2 \ dx# using power rule # - 1/y =-3 x^3 +C# # 1/y = 3 x^3 +C# # y =1/ (3 x^3 +C)# 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 8202 views around the world You can reuse this answer Creative Commons License