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.