Find the values for which y = e^(wx) satisfy the equation

[MarkSA]

Hello,

My question is:

1) Find the values of w for which y = e^(wx) satisfies the equation y + y' = y''

I tried determining what y' and y'' were. I treated w as a number and came up with:
y' = w * e^(wx)
y'' = w^2 * e^(wx)

I then tried to substitute all of them into the equation and do algebra on them. I ended up with (1 + w)/w^2 = 1. If I get rid of the denominator, I end up with w^2 + w^3 = w^2 and then w^3 = 0 which clearly isn't right. Any suggestions?

 

[Deleted member 4993]

Hello,

My question is:

1) Find the values of w for which y = e^(wx) satisfies the equation

y + y' = y''

\(\displaystyle e^{wx}\cdot (w^2 - w - 1) = 0\)

\(\displaystyle (w^2 - w - 1) = 0 .....Now\,\ continue\)

I tried determining what y' and y'' were. I treated w as a number and came up with:
y' = w * e^(wx)
y'' = w^2 * e^(wx)

I then tried to substitute all of them into the equation and do algebra on them. I ended up with (1 + w)/w^2 = 1. If I get rid of the denominator, I end up with w^2 + w^3 = w^2 and then w^3 = 0 which clearly isn't right. Any suggestions?

 

[MarkSA]

\(\displaystyle e^{wx}\cdot (w^2 - w - 1) = 0\)

\(\displaystyle (w^2 - w - 1) = 0 .....Now\,\ continue\)

thanks for the reply. Could you elaborate how you arrived at \(\displaystyle e^{wx}\cdot (w^2 - w - 1) = 0\)?

 

[skeeter]

y = e[sup:3hh8bh0t]wx[/sup:3hh8bh0t]
y' = we[sup:3hh8bh0t]wx[/sup:3hh8bh0t]
y'' = w[sup:3hh8bh0t]2[/sup:3hh8bh0t]e[sup:3hh8bh0t]wx[/sup:3hh8bh0t]

y + y' = y''

e[sup:3hh8bh0t]wx[/sup:3hh8bh0t] + we[sup:3hh8bh0t]wx[/sup:3hh8bh0t] = w[sup:3hh8bh0t]2[/sup:3hh8bh0t]e[sup:3hh8bh0t]wx[/sup:3hh8bh0t]

0 = w[sup:3hh8bh0t]2[/sup:3hh8bh0t]e[sup:3hh8bh0t]wx[/sup:3hh8bh0t] - we[sup:3hh8bh0t]wx[/sup:3hh8bh0t] - e[sup:3hh8bh0t]wx[/sup:3hh8bh0t]

0 = e[sup:3hh8bh0t]wx[/sup:3hh8bh0t](w[sup:3hh8bh0t]2[/sup:3hh8bh0t] - w - 1)