Question #1eb38

Question

1345 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:

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

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

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

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