What is a solution to the differential equation #dy/dx=xy^2# with the particular solution #y(2)=-2/5#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jul 11, 2016 # y = - 2/(x^2 + 1)# Explanation: #dy/dx=xy^2# this is separable #1/y^2 \ dy/dx=x# #int \ 1/y^2 \ dy/dx \ dx =int \ x \ dx# #int \ 1/y^2 \ dy =int \ x \ dx# power rule # - 1/y = x^2/2 + C# #y(2) = - 2/5# # 5/2 = 2 + C implies C = 1/2# # - 1/y = 1/2 (x^2 + 1)# # - 2/y = (x^2 + 1)# # - y/2 = 1/(x^2 + 1)# # y = - 2/(x^2 + 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 29321 views around the world You can reuse this answer Creative Commons License