Y"-28y'+196y=2sin4x-2cos4x

[latex]y''-28y'+196y=2sin4x-2cos4x\\\\1)\; k^2-28k+196=0\\\\(k-14)^2=0\\\\k_1=k_2=14\\\\y_{obshee \; resh.\; odnor.}-=e^{14x}(C_2+C_2x)\\\\2)\; f(x)=e^{0\cdot x}(2sin4x-2cos4x)\; \; \to \; \; \alpha + \beta i=0+4i\ne k_1\; \to \\\\y_{chastn.\; resh.\; neodnor}=Acos4x+Bsin4x\\\\y'=-4asin4x+4Bcos4x\\\\y''=-16Acos4x-16Bsin4x[/latex] [latex]y''-28y'+196y=-16Acos4x-16Bsin4x-\\\\-28(-4Asin4x+4Bcos4x)+Acos4x+Bsin4x=[/latex]                             [latex]=2sin4x-2cos4x[/latex] [latex]cos4x|\, -2=196A-28\cdot 4B-16A\\\\sin4x|\, 2=196B-28\cdot (-4A)-16B\\\\ \left \{ {{180A-112B=-2} \atop {112A+180B=2}} \right. \; \left \{ {{90A-56B=-1} \atop {56A+90B=1}} \right. \; A=-\frac{453544}{282240},\; B=-\frac{8043}{3136}\\\\y_{ch,resh}=-\frac{453544}{282240}cos4x-\frac{8043}{3136}sin4x\\\\3)y_{obsh.resh.neodn}=e^{14x}(C_1+C_2x)-\frac{453544}{282240}cos4x-\frac{8043}{3136}sin4x[/latex] P.S. Проверьте решение системы.