What is a general solution to the differential equation #y'-3y=5#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jun 26, 2016 #= gamma e^ {3x} - 5/3# Explanation: #y'-3y=5# #y'= 3y+5# #(y')/(3y+5) =1# #int \ (dy)/(3y+5) =int \ dx# #1/3ln(3y+5) =x + alpha# #ln(3y+5) =3( x + alpha)# #3y+5 = e^ {3( x + alpha)} = e^ {3x} e^ {3 alpha} = beta e^ {3x}# #y = 1/3 ( beta e^ {3x} - 5)# #= gamma e^ {3x} - 5/3# 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 10279 views around the world You can reuse this answer Creative Commons License