What is a solution to the differential equation #e^xdy/dx+2e^xy=1#?

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Answer 1 · 2016-07-11T20:57:02

# y = e^-x + Ce^{-2x}#

#e^xdy/dx+2e^xy=1#

#dy/dx+2y=e^-x#

this is non-separable, we use an integrating factor (IF)

#IF = e^{int 2 dx}= e^(2x)#

#e^(2x)dy/dx+2e^(2x)y=e^-x * e^(2x)#

#e^(2x)dy/dx+2e^(2x)y=e^x#

#d/dx y e^(2x) = e^x#

# y e^(2x) = int \ e^x \ dx#

# y e^(2x) = e^x + C#

# y = e^-x + Ce^{-2x}#

Answer 2 · 2016-07-11T23:01:35

#y = C_0 e^{-2x}+e^{-x}#

Dividing by #e^x#

#(dy)/(dx)+2y=e^{-x}#

this linear non-homogeneus differential equation has as solution

#y=y_h+y_p#

such that

#(dy_h)/(dx)+2y_h=0#

and

#(dy_p)/(dx)+2y_p=e^{-x}#

For the homogeneus we obtain easily ( variables grouping)

#y_h = C_0 e^{-2x}#

and for the particular, #y_p = e^{-x}# verifies the differential condition.

so the solution is

#y = C_0 e^{-2x}+e^{-x}#