What is a solution to the differential equation dy/dx=xe^y? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Aug 29, 2016 y = ln (1/(C - x^2/2)) Explanation: y' = xe^y e^(-y)y' = x int e^(-y)y' dx =int x dx int e^(-y) dy =int x dx -e^(-y) =x^2/2 + C e^(-y) =C - x^2/2 -y =ln (C - x^2/2) y =- ln (C - x^2/2) y = ln (1/(C - x^2/2)) 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 13594 views around the world You can reuse this answer Creative Commons License