How to you find the general solution of #dy/dx=sin2x#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Douglas K. Nov 27, 2016 Use the separation of variables method. Answer: #y = -1/2cos(2x) + C# Explanation: #dy = sin(2x)dx# #intdy = intsin(2x)dx# #y = -1/2cos(2x) + 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 6470 views around the world You can reuse this answer Creative Commons License