Please help me solve this equation?

Question

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(-1)= 4.

please help with this

Answer 1

You should be able to solve for c by plugging in x= -1 and y = 4 into the first equation.  

Answer 2

First, simply substitute -1 into all values for x in the first formula.

Therefore 4 = ce^(-2*-1) + e^(-(-1))

4 = ce^2 + e^1

4 - e = ce^2

1.2817/ e^2 = c

c = 0.1735

To prove the first part, differentiate y, which becomes

y' = -2ce^(-2x) - e^-x

and so

-2ce^(-2x) - e^-x + 2(ce^(-2x) + e^(-x)) should equal e^-x

-2ce^(-2x) - e^-x + 2ce^(-2x) + 2e^-x

combining like terms, we are left with

e^-x

Done!

Answer 3

I don't understand the second equation. If e represents Euler's (pronounced "oilers") Number, then e is approximately equal to 2.71828182846, so simply substitute this number for e and (as you know) substitute -1 (negative 1) for x. Of course y = 4 when x = -1.

Therefore

4 = c(e^2) + e

Substitute for e and solve for c and you're done!

Have a lot of fun!