# simple question

• May 12th 2010, 04:53 AM
llyold
simple question
hai there, is anyone can show the step-by-step how to solve this equation.please2..

y''+25y = 10 sin pt whre p is positive value

y(0) = 0
y'(0) = 5

tq
• May 12th 2010, 05:04 AM
llyold
the answer should be p = 5

and y(t) = 0.3 sin 5t + t cos 5t..

please anyone show it to me.tq
• May 12th 2010, 05:30 AM
dwsmith
Quote:

Originally Posted by llyold
hai there, is anyone can show the step-by-step how to solve this equation.please2..

y''+25y = 10 sin pt whre p is positive value

y(0) = 0
y'(0) = 5

tq

Solve complement and then solve particular via annihilator method.

$\displaystyle m^2+25=0$ $\displaystyle m=\pm5\mathbf{i}$

$\displaystyle y_c=c_1cos(5t)+c_2sin(5t)$

$\displaystyle (m^2+p^2)(m^2+25)=0$ $\displaystyle m=\pm5\mathbf{i}, \pm p\mathbf{i}$

$\displaystyle y_p=c_3cos(pt)+c_4sin(pt)$

$\displaystyle y'_p=-pc_3sin(pt)+pc_4cos(pt)$

$\displaystyle y''_p=-p^2c_3cos(pt)-p^2c_4sin(pt)$

$\displaystyle -p^2c_3cos(pt)-p^2c_4sin(pt)+25c_3sin(pt)+25c_4cos(pt)=10sin(pt)$

$\displaystyle c_3(-p^2cos(pt)+25sin(pt))+c_4(-p^2sin(pt)+25cos(pt))=10sin(pt)$

I have to get ready to hit the beach so this all I could come up with at the moment.