Question #1eb38

Question

957 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-18T13:14:50

# y = e^(-kx)( x + C )#

#dy/dx + ky = e^(-kx)#

We apply an Integrating Factor (IF): #e^(int k dx) = e^(kx)#

(Note that here the integration constant can be left out as it will cancel when we multiply both sides by the IF.)

Applying the IF to both sides:

#e^(kx)dy/dx +e^(kx) ky = e^(kx)e^(-kx)#

#implies (e^(kx) \ y )' = 1#

#implies e^(kx) \ y = int dx = x + C#

#implies y = e^(-kx)( x + C )#