Question

y''-6y'+8=3x^2+2x+1 помогите найти общее решение

Answer 1

Ответ:

$y''-6y'+8y=3x^2+2x+1\\\\a)\ \ k^2-6k+8=0\ \ ,\ \ k_1=2\ ,\ k_2=4\\\\y_{oo}=C_1e^{2x}+C_2e^{4x}\\\\b)\ \ f(x)=e^{0\cdot x}(3x^2+2x+1)\ ,\ 0\ne 2\ ,\ 0\ne 4\ \to \ \ \ y_{chastn.}=Ax^2+Bx+C\\\\ y_{chastn.}=Ax^2+Bx+C\\ y'_{chastn.}=2Ax+B\\ y''_{chastn.}=2A\\-------------\\y''-6y'+8y=2A-6(2Ax+B)+8(Ax^2+Bx+C)\\\\2A-12Ax-6B+8Ax^2+8Bx+8C=3x^2+2x+1\\\\x^2\ |\ 8A=3\ \ \qquad \qquad \qquad \ \ A=\dfrac{3}{8}\\x\ \ |\ -12A+8B=2\ ,\ \ 4B=6\cdot \dfrac{3}{8}+1\ ,\ \ 4B=\dfrac{9}{4}+1=\dfrac{13}{4}\, ,\ B=\dfrac{13}{16}$

$x^0\ |\ 2A-6B+8C=1\ ,\quad \quad 8C=1-2\cdot \dfrac{13}{16}+6\cdot \dfrac{3}{8}=\dfrac{13}{8}\ ,\ C=\dfrac{13}{64}$

$y_{chastn}=\dfrac{3}{8}\, x^2+\dfrac{13}{16}\, x+\dfrac{13}{64}\\\\\\c)\ \ y_{o.n.}=C_1e^{2x}+C_2e^{4x}+\dfrac{3}{8}\, x^2+\dfrac{13}{16}\, x+\dfrac{13}{64}$