differential equation

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

$\displaystyle y(x) =$

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

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

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

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

i need help applying the condition
• Jul 28th 2007, 10:54 AM
CaptainBlack
Quote:

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

$\displaystyle y(x) =$

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

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

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

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

i need help applying the condition

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

$\displaystyle C_1+C_2=1$

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

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

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

RonL
• Jul 28th 2007, 03:50 PM
topsquark
Quote:

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

$\displaystyle y(x) =$

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

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

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

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

i need help applying the condition

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

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

-Dan
• Jul 28th 2007, 06:25 PM
viet
yeah it was a typo, thanks for pointing that out topsquark
• Jul 29th 2007, 06:39 PM
viet
Quote:

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

$\displaystyle C_1+C_2=1$

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

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

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

RonL

can you explain how to solve that? im pretty much lost.
• Jul 29th 2007, 07:22 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,
$\displaystyle C_1+C_2=5$
$\displaystyle C_1-C_2=1$
Add,
$\displaystyle 2C_1=6\implies C_1=3$
And so on ....

The same idea here.
• Jul 29th 2007, 07:46 PM
CaptainBlack
Quote:

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

$\displaystyle C_1+C_2=1$

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

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

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

RonL

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

$\displaystyle 2~C_1+2~C_2=2$

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):

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

or:

$\displaystyle 7~C_2=12$

so:

$\displaystyle C_2=12/7$

Now substitute this back into the first equation to get:

$\displaystyle C_1=-5/7$

RonL