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It is easy to check that for any value of $c$, the function $$y = ce^{-2x} + e^{-x}$$

is solution of equation $$y' + 2y = e^{-x}.$$ Find the value of $c$ for which the solution satisfies the initial condition $y(-5)= 6$

I start out by making it $y' =-2y+ e^{-x}$

this gives me $df/dy = -2y$ and $df/dx = e^{-x}$

I'm stuck here not sure what to do next.

Moo
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David Scidmore
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  • @DavidScidmore : You actually want to use $y = ce^{-2x} + e^{-x}$ to find $c$. You used the wrong equation. – Moo Jan 23 '16 at 21:41

1 Answers1

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You already have the formula $y = c e^{-2x} + e^{-x}$. All you need to do is find $c$ such that $y(-5) = 6$. So let's plug in $x = -5$ and the corresponding $y = 6$: $$ 6 = c e^{10} + e^{5} \iff c e^{-10} = 6 - e^{5} \iff c = e^{-10}(6 - e^{5}) \iff c = 6 e^{-10} - e^{-5}). $$ (The numerical value of $c$ is approximately $-0.000646555$.)

If you were to have the ODE $y' + 2y = e^{-x}$, then you would first solve to via the complimentary/auxiliary function and particular solution method to get $y = c e^{-2x} + e^{-x}$ for a constant $c$, and then plug in your initial condition to get $c$, as I've done above.

Sam OT
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  • What I've done now is.

    6=ce^10+e^5

    6 = e^5 (ce^2 + 1) 6/e^5 = ce^2+1

    (6/e^5)-1 = ce^2

    (6/e^7)-1/e^2 = c

     – David Scidmore Jan 23 '16 at 22:05
  • I've just corrected my answer -- I put in $x = +5$ instead of $x = -5%, whoops! You've made a mistake above, though: remember that $e^a e^b = e^{a+b} \neq e^{ab}$. If you see my answer, I've added a few intermediary steps to show the details. :) – Sam OT Jan 23 '16 at 22:16