What is a solution to the differential equation #xdy/dx=1/y#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jul 7, 2016 #y = pm sqrt{ ln x^2 + C}# Explanation: this is separable #x dy/dx=1/y# #y dy/dx=1/x# #int y dy=int 1/x dx# #y^2 /2 = ln x + C# #y^2 = 2 ln x + C = ln x^2 + C# #y = pm sqrt{ ln x^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 6603 views around the world You can reuse this answer Creative Commons License