# differential equation

• Jul 28th 2007, 11:44 AM
viet
differential equation
solve $y'' - 3y' - 10 y = 0; \quad y = 1, y'=10 \;\textnormal{at}\; x = 0$

$y(x) =$

heres what i did:
$r^2-3r-10 = 0$

$(r-2) (r-5) = 0$
roots are : -2, 5

General solution: $y = C_1 e^{-2x}+C_2 e^{5x}$

$y' = -2C_1 e^{-2x}+5C_2 e^{5x}$

i need help applying the condition
• Jul 28th 2007, 11:54 AM
CaptainBlack
Quote:

Originally Posted by viet
solve $y'' - 3y' - 10 y = 0; \quad y = 1, y'=10 \;\textnormal{at}\; x = 0$

$y(x) =$

heres what i did:
$r^2-3r-10 = 0$

$(r-2) (r-5) = 0$
roots are : -2, 5

General solution: $y = C_1 e^{-2x}+C_2 e^{5x}$

$y' = -2C_1 e^{-2x}+5C_2 e^{5x}$

i need help applying the condition

The initial condition $y(0)=1$ gives:

$
C_1+C_2=1
$

and the condition $y'(0)=10$ gives:

$
-2~C_1+5~C_2=10
$

So now you have a pair of simultaneous equations for $C_1$ and $C_2$ to solve.

RonL
• Jul 28th 2007, 04:50 PM
topsquark
Quote:

Originally Posted by viet
solve $y'' - 3y' - 10 y = 0; \quad y = 1, y'=10 \;\textnormal{at}\; x = 0$

$y(x) =$

heres what i did:
$r^2-3r-10 = 0$

$(r-2) (r-5) = 0$
roots are : -2, 5

General solution: $y = C_1 e^{-2x}+C_2 e^{5x}$

$y' = -2C_1 e^{-2x}+5C_2 e^{5x}$

i need help applying the condition

For the record, the characteristic equation factors as
$(r + 2)(r - 5) = 0$

Given that you have the solution to the differential equation correct I would guess this is merely a typo?

-Dan
• Jul 28th 2007, 07:25 PM
viet
yeah it was a typo, thanks for pointing that out topsquark
• Jul 29th 2007, 07:39 PM
viet
Quote:

Originally Posted by CaptainBlack
The initial condition $y(0)=1$ gives:

$
C_1+C_2=1
$

and the condition $y'(0)=10$ gives:

$
-2~C_1+5~C_2=10
$

So now you have a pair of simultaneous equations for $C_1$ and $C_2$ to solve.

RonL

can you explain how to solve that? im pretty much lost.
• Jul 29th 2007, 08:22 PM
ThePerfectHacker
Quote:

Originally Posted by viet
can you explain how to solve that? im pretty much lost.

This is just a linear system.

Say,
$C_1+C_2=5$
$C_1-C_2=1$
$2C_1=6\implies C_1=3$
And so on ....

The same idea here.
• Jul 29th 2007, 08:46 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
The initial condition $y(0)=1$ gives:

$
C_1+C_2=1
$

and the condition $y'(0)=10$ gives:

$
-2~C_1+5~C_2=10
$

So now you have a pair of simultaneous equations for $C_1$ and $C_2$ to solve.

RonL

To solve these multiply the firs equation by 2 to get:

$
2~C_1+2~C_2=2
$

Now add the third equation to this (that is add the left hand sides, which
sum is equal to the sum of the right hand sides):

$
2~C_1+2~C_2-2~C_1+5~C_2=2+10
$

or:

$
7~C_2=12
$

so:

$
C_2=12/7
$

Now substitute this back into the first equation to get:

$
C_1=-5/7
$

RonL