Question

Решить дифференциальное уравнение: y"+9y'=cos(2x)

Answer 1

$y''+9y'=cos2x\\\\a)\ \ k^2+9k=0\ \ ,\ \ k(k+9)=0\ \ ,\ \ k_1=0\ ,\ \ k_2=-9\\\\y_{o.odn.}=C_1+C_2e^{-9x}\\\\b)\ \ f(x)=cos2x=e^{0x}\cdot (1\cdot cos2x+0\cdot sin2x)\ \ ,\ \alpha +\beta i=0+2i\ne k_{1,2}\\\\\widetilde{y}=Acos2x+Bsin2x\\\\\widetilde{y}'=-2Asin2x+2Bcos2x\\\\\widetilde{y}''=-4Acos2x-4Bsin2x\\-----------------------------\\y''+9y'=-4Acos2x-4Bsin2x+9Acos2x+9Bsin2x=cos2x\\\\cos2x\ |\ -4A+9A=1\ \ ,\ \ \ 5A=1\ \ ,\ \ A=\dfrac{1}{5}\\sin2x\ |\ -4B+9B=0\ \ ,\ \ \ 5B=0\ \ ,\ \ B=0$

$\widetilde{y}=\dfrac{1}{5}\, cos2x\\\\c)\ \ \ y_{obsh.neodn.}=C_1+C_2\, e^{-9x}+\dfrac{1}{5}\, cos2x$