# Basic differential equations help

• Feb 7th 2010, 05:02 PM
lordpikachu
Basic differential equations help
find the general solution for y''[x] = -y'[x]

the answer is y=c1e^(tx1) + c2e^(tx2) where c1, c2, x1, x2 are separate variables/constants...
i don't understand..i keep getting that y=ce^-x but that comes nowhere close to what the real answer is?

If anyone can help me, I would greatly appreciate it (ie worship them forever)

Thanks

ps. should this go in the diffeq forum? i wasn't sure because this is basic diffeq you learn in calc...
• Feb 7th 2010, 06:31 PM
tonio
Quote:

Originally Posted by lordpikachu
find the general solution for y''[x] = -y'[x]

the answer is y=c1e^(tx1) + c2e^(tx2) where c1, c2, x1, x2 are separate variables/constants...
i don't understand..i keep getting that y=ce^-x but that comes nowhere close to what the real answer is?

If anyone can help me, I would greatly appreciate it (ie worship them forever)

Thanks

ps. should this go in the diffeq forum? i wasn't sure because this is basic diffeq you learn in calc...

You are correct: observe that $y''=-y\Longrightarrow \frac{y''}{y'}=-1\Longrightarrow \ln y'=-x+C\Longrightarrow y'=Ke^{-x}\Longrightarrow y=Ke^{-x}$ , integrating twice.

Tonio
• Feb 7th 2010, 06:40 PM
vince
Quote:

Originally Posted by lordpikachu
find the general solution for y''[x] = -y'[x]

the answer is y=c1e^(tx1) + c2e^(tx2) where c1, c2, x1, x2 are separate variables/constants...
i don't understand..i keep getting that y=ce^-x but that comes nowhere close to what the real answer is?

If anyone can help me, I would greatly appreciate it (ie worship them forever)

Thanks

ps. should this go in the diffeq forum? i wasn't sure because this is basic diffeq you learn in calc...

set up the associated characteristic polynomial and solve for the roots. those roots are what appear in the exponents of the exponentials above.
• Feb 7th 2010, 06:52 PM
vince
Quote:

Originally Posted by tonio
You are correct: observe that $y''=-y\Longrightarrow \frac{y''}{y'}=-1\Longrightarrow \ln y'=-x+C\Longrightarrow y'=Ke^{-x}\Longrightarrow y=Ke^{-x}$ , integrating twice.

Tonio

This works. I get $y=c1 + c2e^{-x}$
• Feb 7th 2010, 08:19 PM
lordpikachu
OHHH i get it
Oh wow thanks!! I LOVE YOU GUYS. // thanks a million times.
• Feb 8th 2010, 05:29 AM
HallsofIvy
Quote:

Originally Posted by lordpikachu
find the general solution for y''[x] = -y'[x]

the answer is y=c1e^(tx1) + c2e^(tx2) where c1, c2, x1, x2 are separate variables/constants...
i don't understand..i keep getting that y=ce^-x but that comes nowhere close to what the real answer is?

The general solution is, as you have already been told, $y= c_1+ c_2e^{-t}$ (it is your choice to use t or x as the independent variable since it is not given in the equation). My point is that this is exactly of the form $c_1e^{tx_1}+ c_2e^{tx_2}$ with $x_1= 0$ and $x_2= -1$.

Quote:

If anyone can help me, I would greatly appreciate it (ie worship them forever)
We will forward you an address to send your tithes to. You will have to purchase your own white robes.

Quote:

Thanks

ps. should this go in the diffeq forum? i wasn't sure because this is basic diffeq you learn in calc...