# Thread: 2ODE Linear with one constant?

1. ## 2ODE Linear with one constant?

$y^{\prime \prime} + 25 y = 0$

Initial Conditions are:
$y(0)=-1,\quad y(\frac{\pi}{10})=-2$

So, My roots are $0\pm10i$

Its in the form P+- qi
Does the fact that my P=0 change anything? Or should I just roll with it?

Anyways, heres what I did

$Y=ACos(5x)+Bsin(5x))$
$Y'=-A5sin(5x)+B5Cos(5x)$

So, when Y(0)=-1
We get $A=-1$

When $Y'(\frac{\pi}{10})=-2$
we get:
$-2=-A5$ which is just a false statement.
Im not able to isolate "B" given the initial conditions. What am I doing wrong?

2. I'm pretty sure that the roots aren't $\displaystyle 0 \pm 10i$...

3. Ur right, Its 0 +- 5i

Thats the value I actually used, I just copied it here wrong.
But still, my problem persists!

4. Originally Posted by Vamz
$y^{\prime \prime} + 25 y = 0$

Initial Conditions are:
$y(0)=-1,\quad y(\frac{\pi}{10})=-2$

So, My roots are $0\pm10i$

Its in the form P+- qi
Does the fact that my P=0 change anything? Or should I just roll with it?

Anyways, heres what I did

$Y=ACos(5x)+Bsin(5x))$
$Y'=-A5sin(5x)+B5Cos(5x)$

So, when Y(0)=-1
We get $A=-1$

When $Y'(\frac{\pi}{10})=-2$
we get:
$-2=-A5$ which is just a false statement.
Im not able to isolate "B" given the initial conditions. What am I doing wrong?
You differentiated y(x) for the condition y(pi/10) which you didnt need to.

ie just use $Y=ACos(5x)+Bsin(5x))$ for both conditions