# Solve Initial Value Problem given General Solution.

• March 1st 2011, 05:07 PM
Jeonsah
Solve Initial Value Problem given General Solution.
Im having a little trouble with these problems. Im sure its pretty simple. Heres one from my book:

Given the family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem.

$y = C1e^x + C2e^-^x, (-\infty, \infty)$
$y'' - y = 0, y(0) = 0, y'(0) = 1$

Can someone show me how to do this problem? I am just confused one how to work it. Im sure its pretty simple but I cant seem to figure it out.

Thank you!
• March 1st 2011, 07:25 PM
Ackbeet
Your goal is to plug in the initial conditions $y(0)=0, y'(0)=1,$ and solve for $C_{1},C_{2}.$ So what two equations do you get by plugging in the initial conditions into your general solution?
• March 1st 2011, 09:43 PM
Jeonsah
plugging in $y(0) = 0$
$C1 + C2 = 0$
plugging in $y'(0) = 1$
$C1 - C2 = 1$
• March 2nd 2011, 02:41 AM
Ackbeet
Good so far! What next?
• March 2nd 2011, 10:33 AM
Jeonsah
• March 2nd 2011, 10:36 AM
Ackbeet
Sure. What does that give you?
• March 2nd 2011, 10:39 AM
Jeonsah
C1 = 1/2
C2 = -1/2

Then
$y = .5e^x - .5e^-^x$

thats it right?
• March 2nd 2011, 10:42 AM
Ackbeet
This is the beautiful thing about differential equations: it's quite straight-forward to check your own answer (hardly more difficult than differentiating, and quite a bit easier than integrating). So I'm going to ping this question right back at you:

1. Does your solution satisfy the DE?
2. Does your solution satisfy the initial conditions?

If so, I'd say you've got it.

You should get into the habit of doing two things to every solution to every DE you ever solve: 1. Find the interval of validity. 2. Check your solution by differentiation.
• March 2nd 2011, 10:46 AM
Jeonsah
Thank you!

the reason I wasnt sure what quite to do is that my teacher crammed 3 sections into an hour since we are behind. Thanks for the help tho, i appreciate it!
• March 2nd 2011, 10:47 AM
Ackbeet
You're welcome!