How do you find the general solution to #dy/dx=(2x)/e^(2y)#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jul 24, 2016 # y = 1/2 ln (2x^2 + C)# Explanation: #dy/dx=(2x)/e^(2y)# this is separable #e^(2y)dy/dx=2x# #int \ e^(2y)dy/dx \ dx=int \ 2x \ dx# #1/2 e^(2y) = x^2 + C# # e^(2y) = 2x^2 + C# # 2y = ln (2x^2 + C)# # y = 1/2 ln (2x^2 + 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 10603 views around the world You can reuse this answer Creative Commons License