Question

Рассчитать коши методом неопределенных коэффициентов y''-2y'+y=-12cos2x-9sin2x. y(0)=-2; y'(0)=0

Answer 1

$y''-2y'+y=-12cos2x-9sin2x\; ,\; \; y(0)=-2\; ,\; \; y'(0)=0\\\\1)\; \; k^2-2k+1=0\; ,\; \; (k-1)^2=0\; ,\; \; k_1=k_2=1\\\\y_{obsh.resh.odnor.}=e^{x}\cdot (C_1x+C_2)\\\\2)\; \; f(x)=e^{0\cdot x}\cdot (-12cos2x-9sin2x)\; ,\; \; 0\pm 2i\ne k_{1,2}\\\\y_{chastn.resh.neodn.}=Acos2x+Bsin2x\\\\y'=-2Asin2x+2Bcos2x\\\\y''=-4Acos2x-4Bsin2x\\------------------\\y''-2y'+y=-4Acos2x-4Bsin2x+4Asin2x-4Bcos2x+\\\\+Acos2x+Bsin2x=cos2x(-3A-4B)+sin2x(-3B+4A)\; ;\\\\(-3A-4B)cos2x+(-3B+4A)sin2x=-12cos2x-9sin2x$

$\left \{ {{-3A-4B=-12} \atop {4A-3B=-9}} \right. \oplus \left \{ {{A-7B=-21} \atop {4A-3B=-9}} \right. \; \left \{ {{A=7B-21} \atop {28B-84-3B=-9}} \right. \; \left \{ {{A=0} \atop {B=3}} \right. \\\\y_{chast.resh.ntodn.}=3\cdot sin2x\\\\3)\; \; y_{obshee\; neodn.}=e^{x}\cdot (C_1x+C_2)+3\cdot sin2x\\\\4)\; \; y(0)=-2\; ,\; \; -2=C_2+3\cdot sin0\; ,\; \; C_2=-2$

$y'(0)=0\; ,\; \; y_{obsh.neodn.}'=e^{x}(C_1x+C_2)+e^{x}\cdot C_1+6\cdot cos2x\\\\0=C_2+C_1+6\cdot cos0\\\\0=-2+C_1+6\cdot \frac{\pi}{2}\; ,\; \; C_1=2-3\pi$

$4)\; \; y_{chastn.neodnor.}=e^{x}\cdot ((2-3\pi )x-2)+3\cdot sin2x$