1. ## 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

2. 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

3. 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

4. yeah it was a typo, thanks for pointing that out topsquark

5. 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.

6. 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.

7. 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