Question

Can someone help me to solve this differential eqation please.

y''+9y=cosx

(i got solution to homogeneuos equation: c1*cos(3x)+c2*sin(3x) )
i have tried to find particular with variation of parameters, with wronskian...i have so weird integral :(

Answer 1

try going to wolframalpha.com

Answer 2

have u tried undetermined coeffecients?

Answer 3

\[y_p=Acosx + Bsinx\]\[y_p'=-Asinx+Bcosx\]\[y_p''=-Acosx-Bsinx\]substitute in equation
\[-Acosx-Bsinx+9(Acosx+Bsinx)=cosx\]\[-Acosx-Bsinx+9Acosx +9Bsinx=cosx\]\[8Acosx+8Bsinx=cosx\]\[8A=1\]\[A=\frac{1}{8}\]\[8B=0\]\[B=0\]
so \[y_p=\frac{1}{8}cosx\]\[y=y_h + y_p\]\[=c_1\cos3x+c_2\sin3x +\frac{1}{8}cosx\]