What is a general solution to the differential equation #y'=2+2x^2+y+x^2y#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jul 24, 2016 #y = Ce^(x + x^3/3) - 2# Explanation: #y'=2+2x^2+y+x^2y# this is separable, #y'=2(1+x^2)+y(1+x^2)# #y'=(2 + y)(1+x^2)# #1/(2+y) y'=(1+x^2)# #int \ 1/(2+y) y' \ dx =int \ (1+x^2) \ dx# #int \ d/dx( ln(2+y) ) \ dx =int \ (1+x^2) \ dx# #ln(2+y) =x + x^3/3 + C# #2+y = e^(x + x^3/3 + C)# #2+y = Ce^(x + x^3/3) # #y = Ce^(x + x^3/3) - 2# 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 6050 views around the world You can reuse this answer Creative Commons License