What is a solution to the differential equation y'-y=5? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Aug 7, 2016 y= C e^x - 5 Explanation: separate it! y' - y = 5 y' = y + 5 1/(y+5) y' = 1 int 1/(y+5) y' \ dx=int \dx int d/dx (int 1/(y+5) \dy) \ dx=int \dx int 1/(y+5) \dy=int \dx ln(y+5) =x + C y+5 =e^(x + C) = C e^x y= C e^x - 5 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 18818 views around the world You can reuse this answer Creative Commons License