What is a solution to the differential equation dy/dx=2y-1? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Cesareo R. Aug 23, 2016 y = (C_2 e^{2x}+1)/2 Explanation: Substituting z = 2y-1 in dy/dx=2y-1 and considering dz/(dx) = 2 dy/(dx) we have 1/2 dz/dx = z. Grouping variables dz/z = 2 dx and integrating log_e z = 2 x + C_1 or equivalently z = C_2e^{2x} = 2 y -1 and finally y = (C_2 e^{2x}+1)/2 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 7008 views around the world You can reuse this answer Creative Commons License