How to you find the general solution of #dy/dx=x/y#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Roy E. Jan 13, 2017 #x^2-y^2=c# Explanation: #dy/{dx}=x/y# #ydy=xdx# by exploiting the notation (separation) #int ydy=int xdx# further exploiting the notation #1/2y^2=1/2x^2+d# #y^2=x^2+2d# #x^2-y^2=-2d# #x^2-y^2=c# where #c=-2d# Depending on whether #c# is positive, negative or zero you get a hyperbola open to the #x#-axis, open to the #y#=axis, or a pair of straight lines through the origin. 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 19501 views around the world You can reuse this answer Creative Commons License