Question

Помогите решить уравнение
y''-y'-6y=3x^2-x+4

Answer 1

$y''-y'-6y=3x^2-x+4 \\\\ y''-y'-6y=0 \\ y=e^{kx} \\ k^2e^{kx}-ke^{kx}-6e^{kx}=0 \\ e^{kx}(k^2-k-6)=0 \\k^2-k-6=0 \\ (k-3)(k+2)=0 \\ y=C_1e^{3x}+C_2e^{-2x} \\\\ y^*=Ax^2+Bx+C \\ y^*'=2Ax+B \\ y^*''=2A \\\\ 2A-2Ax-B-6Ax^2-6Bx-6C=3x^2-x+4 \\ -6A=3; A=-0.5 \\ -2A-6B=-1; B=\frac{1}{3} \\2A-B-6C=4;-6C=4+1+\frac{1}{3}; C=-\frac{8}{9} \\\\y^*=-0.5x^2+\frac{1}{3}x-\frac{8}{9} \\\\ Y=C_1e^{3x}+C_2e^{-2x} -0.5x^2+\frac{1}{3}x-\frac{8}{9}$

Answer 2

$y''-y'-6y=3x^2-x+4\\\lambda^2-\lambda-6=0\\\lambda_1=-2\ \ \ \ \ \ \lambda_2=3\\Y=C_1e^{-2x}+C_2e^{3x}\\\hat{y}=Ax^2+Bx+C\\\hat{y}'=2Ax+B\\\hat{y}''=2A\\2A-2Ax-B-6Ax^2-6Bx-6C=3x^2-x+4\\x^2|-6A=3;A=-\frac{1}{2}\\x|-2A-6B=-1;B=\frac{1}{3}\\x^0|2A-B-6C=4;C=-\frac{8}{9}\\\hat{y}=-\frac{x^2}{2}+\frac{x}{3}-\frac{8}{9}\\y=Y+\hat{y}=C_1e^{-2x}+C_2e^{3x}-\frac{x^2}{2}+\frac{x}{3}-\frac{8}{9}$