How do you find the general solution to dy/dx+e^(x+y)=0? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jul 24, 2016 y = ln ( 1/(e^x + C) ) Explanation: dy/dx+e^(x+y)=0 this is separable dy/dx= - e^(x+y) dy/dx= - e^x e^y e^(-y) dy/dx= - e^x int\ e^(-y) dy/dx \ dx=int - e^x \ dx int\ e^(-y) \ dy=- int e^x \ dx -e^(-y) =- e^x + C e^(-y) = e^x + C e^(y) = 1/(e^x + C) y = ln ( 1/(e^x + 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 9116 views around the world You can reuse this answer Creative Commons License