Asked by [email protected]
find the values of k for y= e^kx satisfies the equation y+y'=y"
Answers
Answered by
tammy
hey bot, that is impressive, you are correct
Answered by
Bot
Substitute in y= e^kx:
y' = ke^kx
y" = k^2e^kx
Now we can write the equation as:
e^kx + ke^kx = k^2e^kx
Combine like terms:
(1+k-k^2) e^kx = 0
For this to be true for all x, we must have:
1+k-k^2 = 0
This is a quadratic equation in k, so we can solve using the quadratic formula:
k = (-1 ± √5)/2
Therefore, the values of k that satisfy the equation y= e^kx and y+y'=y" are:
k = (-1 + √5)/2 and k = (-1 - √5)/2.
y' = ke^kx
y" = k^2e^kx
Now we can write the equation as:
e^kx + ke^kx = k^2e^kx
Combine like terms:
(1+k-k^2) e^kx = 0
For this to be true for all x, we must have:
1+k-k^2 = 0
This is a quadratic equation in k, so we can solve using the quadratic formula:
k = (-1 ± √5)/2
Therefore, the values of k that satisfy the equation y= e^kx and y+y'=y" are:
k = (-1 + √5)/2 and k = (-1 - √5)/2.
Answered by
Bot
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