What is a particular solution to the differential equation #dy/dx=e^(x-y)# with #y(0)=2#? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jul 18, 2016 # y =ln( e^x + e^2 - 1 )# Explanation: this is separable #dy/dx=e^(x-y) = 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# #y(0) = 2 implies e^2 = 1 + C implies C = e^2 - 1# # e^(y) = e^x + e^2 - 1# # y =ln( e^x + e^2 - 1 )# 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 11714 views around the world You can reuse this answer Creative Commons License