Question Number 103788 by bemath last updated on 17/Jul/20

y′′−y = cot x

Answered by mathmax by abdo last updated on 18/Jul/20

y^(′′) −y =((cosx)/(sinx))  h→r^2 −1=0 ⇒r =+^− 1 ⇒y_h =a e^x  +be^(−x)  =au_1  +bu_2   W(u_1 ,u_2 ) = determinant (((e^x           e^(−x) )),((e^x          −e^(−x) )))=−2≠0  W_1 = determinant (((o         e^(−x) )),((((cosx)/(sinx))     −e^(−x) )))=−((e^(−x) cosx)/(sinx))  W_2 = determinant (((e^x             0)),((e^x           ((cosx)/(sinx)))))=((e^x  cosx)/(sinx))  v_1 =∫ (w_1 /w)dx =∫  ((e^(−x) cosx)/(2sinx)) dx  v_2 =∫ (w_2 /w)dx =−(1/2)∫ ((e^x cosx)/(sinx))dx  ⇒y_p =u_1 v_1  +u_2 v_2  =e^(x )  ∫  ((e^(−x)  cosx)/(2sinx))dx −e^(−x) ∫ ((e^x cosx)/(2sinx))dx  the general solution is y =y_p  +y_h