Question

Найти общее решение дифференциального уравнения:
y"-9y=3x^2+4x
y"+y'=(16x+24)e^x

Answer 1

$\displaystyle y''-9y=3x^2+4x\\\lambda^2-9=0\\\lambda_{1,2}=^+_-3\\Y=C_1e^{-3x}+C_2e^{3x}\\\hat{y}=Ax^2+Bx+C\\\hat{y}'=2Ax+B\\\hat{y}''=2A\\2A-9Ax^2-9Bx-9C=3x^2+4x\\x^2|-9A=3=\ \textgreater \ A=-\frac{1}{3}\\x|-9B=4=\ \textgreater \ B=-\frac{4}{9}\\x^0|2A-9C=0=\ \textgreater \ C=-\frac{2}{27}\\\hat{y}=-\frac{x^2}{3}-\frac{4x}{9}-\frac{2}{27}\\y=Y+\hat{y}=C_1e^{-3x}+C_2e^{3x}-\frac{x^2}{3}-\frac{4x}{9}-\frac{2}{27}$

$y''+y'=(16x+24)e^x\\\lambda^2+\lambda=0\\\lambda_1=0;\lambda_2=-1\\Y=C_1e^{-x}+C_2\\\hat{y}=e^x(Ax+B)\\\hat{y}'=e^x(Ax+A+B)\\y''=e^x(Ax+2A+B)\\2Ax+3A+2B=16x+24\\x|A=8\\x^0|3A+2B=24=\ \textgreater \ B=0\\\hat{y}=8xe^{x}\\y=Y+\hat{y}=C_1e^{-x}+C_2+8xe^x$