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

2 Answers
Jul 11, 2016

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

Explanation:

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}

Jul 11, 2016

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

Explanation:

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}