# Linear Second Order DE

• Mar 15th 2011, 03:51 AM
anthony162
Linear Second Order DE
The initial question was :

y''-10y'+61y=0

i've got the general solution which is e^5x(A cos(6x)+b sin(6x))

given that the initial conditions were y(0)=0 and y'(0)=-9,find A and B

after i calculated shudn't A =0?it seems i was wrong?

• Mar 15th 2011, 04:04 AM
Ted
A=0 is correct.

What makes you think it is wrong?
• Mar 15th 2011, 04:10 AM
Prove It
$\displaystyle A = 0$ is correct.
• Mar 15th 2011, 04:10 AM
anthony162
then how would the general solution be like?as in y(x)=?
• Mar 15th 2011, 04:11 AM
Prove It
Differentiate it and substitute your other condition to solve for $\displaystyle B$.
• Mar 15th 2011, 04:18 AM
anthony162
I've got B= 54

thus my equation should be 'exp(5*x)*(54*sin(6*x))'..but somehow its wrong
• Mar 15th 2011, 04:23 AM
Prove It
I don't know how you can get $\displaystyle B = 54$...

$\displaystyle y'(x) = 5Be^{5x}\sin{(6x)} + 6Be^{5x}\cos{(6x)}$, and since $\displaystyle y'(0) = -9$

$\displaystyle -9 = 5Be^{5\cdot 0}\sin{(6\cdot 0)} + 6Be^{5\cdot 0}\cos{(6\cdot 0)}$

$\displaystyle -9 = 6B$

$\displaystyle B = -\frac{9}{6} = -\frac{3}{2}$...
• Mar 15th 2011, 04:32 AM
anthony162
Ahh..i integrated it wrongly!!Thanks alot!
• Mar 15th 2011, 05:38 PM
Prove It
Quote:

Originally Posted by topsquark
You made a slight error. A = 0, so the sine term should not be there. It worked because sin(6*0) = 0.

-Dan

You need to use the product rule to differentiate $\displaystyle y(x) = Be^{5x}\sin{(6x)}$.

$\displaystyle u(x) = Be^{5x} \implies u'(x) = 5Be^{5x}$ and $\displaystyle v(x) = \sin{(6x)} \implies v'(x) = 6\cos{(6x)}$.

Therefore $\displaystyle y'(x) = u'(x)v(x) + u(x)v'(x)$

$\displaystyle = 5Be^{5x}\sin{(6x)} + 6Be^{5x}\cos{(6x)}$.

I did not make a mistake.
• Mar 15th 2011, 06:17 PM
topsquark
Quote:

Originally Posted by Prove It
I did not make a mistake.

I....ummmm...I was just testing you. Yeah, that's it. Testing. (crawls away into a dark corner.)

-Dan