Question

find the general solution
\[y'-8 y+18=0\qquad y(2)=1\]

Answer 1

what's qquad

Answer 2

\[y'-8 y+18=0\]

Answer 3

forget it

Answer 4

no idea what quad is

Answer 5

nah, it just said it by your question in my browser

Answer 6

really its an init val prob, but i can do that part.

Answer 7

i didnt mean to copy the y(2) = 1

Answer 8

what i got is \[ln(8y-18) = 8t + C\]

Answer 9

but with the init value it gives an illegal ln value, ln(-10)

Answer 10

\[y'-8y = -18 \Rightarrow e^{-8t}y'-8e^{-8t}y = -18e^{-8t}\]
\[\Rightarrow (ye^{-8t})' = -18e^{-8t} \Rightarrow ye^{-8t} = \frac{-18}{-8}e^{-8t}+C\]
\[y = \frac{9}{4}+Ce^{8t}\]

Answer 11

The answer I got is y=9/4 - (5e^(8x))/(4e^(16))

Answer 12

Now you use the initial value to find out C

Answer 13

If there are any steps you need explained in more detail please ask :)

Answer 14

Bingo. Thanks Dave

Answer 15

No problem, you've helped me in the past ;-)

Answer 16

I had something close but not quite:
\[y(x)=(-5*e^{8(t+2)}+9)/4\]

Answer 17

yeah, instead of the (t+2) it should be minus 2 since there is division there not multiplication