Question

y"+3y'-4y=9x^2-4x+7 найти решение дифференциального уравнения 2-го порядка​

Answer 1

$y''+3y'-4y=9x^2-4x+7\\\\1)\ \ k^2+3k-4=0\ \ ,\ \ k_1=-4\ ,\ k_2=1\\\\y_{oo}=C_1\, e^{-4x}+C_2\, e^{x}\\\\2)\ \ f(x)=9x^2-4x+7\ \ \Rightarrow \ \ y_{chastn.}=Ax^2+Bx+C\\\\y_{chastn.}=Ax^2+Bx+C\\\\y'_{chastn.}=2Ax+B\\\\y''_{chastn.}=2A\\-------------------------\\y''+3y'-4y=2A+3(2Ax+B)-4(Ax^2+Bx+C)\\\\-4Ax^2+(6A-4B)\, x+(2A+3B-4C)=9x^2-4x+7\ \ \ \ \Rightarrow$

$x^2\ |\ -4A=9\ \qquad \ \ \ \ ,\ \ \ \ \ \ A=-\dfrac{9}{4}\ ,\\\\x\ \ |\ 6A-4B=-4\ \ \ \ \ ,\ \ \ \ 4B=6A+4=-\dfrac{6\cdot 9}{4}+4=-\dfrac{19}{2}\ ,\ B=-\dfrac{19}{8}\\\\x^0\ |\ 2A+3B-4C=7\ \ ,4C=2A+3B-7=\dfrac{-18}{4}-\dfrac{57}{8}+7=-\dfrac{37}{8}\ ,\ C=-\dfrac{37}{32}\\\\\\y_{chastn.}=-\dfrac{9}{4}\, x^2-\dfrac{19}{8}\, x-\dfrac{37}{32}\\\\\\3)\ \ y=y_{oo}+y_{chastn.}=C_1\, e^{-4x}+C_2\, e^{x}-\dfrac{9}{4}\, x^2-\dfrac{19}{8}\, x-\dfrac{37}{32}$