Question

Y"+4y'+5y=5x^2-32x+5 срочноооо

Answer 1

$y''+4y'+5y=5x^2-32x+5\\\lambda^2+4\lambda+5=0\\\lambda_{1,2}=-2^+_-i\\Y=e^{-2x}(C_1cosx+C_2sinx)\\\hat{y}=Ax^2+Bx+C\\\hat{y}'=2Ax+B\\\hat{y}''=2A\\2A+8Ax+4B+5Ax^2+5Bx+5C=5x^2-32x+5\\x^2|5A=5=\ \textgreater \ A=1\\x|8A+5B=-32=\ \textgreater \ B=-8\\x^0|2A+4B+5C=5=\ \textgreater \ C=7\\Y=x^2-8x+7\\y=Y+\hat{y}=e^{-2x}(C_1cosx+C_2sinx)+x^2-8x+7$