1. 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

2. the answer should be p = 5

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

please anyone show it to me.tq

3. 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.

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

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

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

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

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

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

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

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

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