y = c1e^x + c2e^−x is a family of solutions for the DE y′′ − y = 0. Find a
solution with initial conditions: y(1) = 0, and y′(1) = e.

Question

y = c1e^x + c2e^−x is a family of solutions for the DE y′′ − y = 0. Find a
solution with initial conditions: y(1) = 0, and y′(1) = e.

anonymous · Answer

how do i go about doing this

anonymous · Answer

i took intergrals 2 times
and evaluated the family of equations

anonymous · Answer

but i could not get 0 or e

anonymous · Answer

y(1)=c1e^(1)+c2e^(-1)=0   ---> Eq. (1)
y'(1)=c1e^1-c2e^(-1)=e     ---> Eq. (2)

solve for c1 & c2:

Eq.(1)+Eq. (2)
2c1 e = e
c1 = 1/2, plugin to either Eq (1) or Eq. (2), solve for c2

Eq. (1):

(1/2)e+c2 e^(-1) =0
c2=-(1/2)e^2

anonymous · Answer

Do u hve any question on the solution here?

anonymous · Answer

i mean it gets the right answer but i think i am messing up with the integrarion of ce^x

anonymous · Answer

hmm you dont integrate here just solve for c1 and c2 first

anonymous · Answer

i guess its fine then

anonymous · Answer

y = c1e^x + c2e^−x
y =(1/2)e^x + (-e^2/2)e^−x  solution to DE

anonymous · Answer

ok good luck now, have fun solving ..lol