найти общее решение дифференциального уравнения)) Y’’-9y’+20y=x^2e^4x;
y”-y’=cos2x+sin2x

Ответ:

КариM

13.12.2020 15:12

$1)\ \ y''-9y'+20y=x^2e^{4x}\\\\a)\ \ k^2-9k+20=0\ \ ,\ \ k_1=4\ ,\ k_2=5\\\\y_{oo}=C_1e^{4x}+C_2e^{5x}\\\\b)\ \ f(x)=x^2e^{4x}\ \ ,\ \ \alpha =4=k_1\ \ \to \ \ r=1\ \ ,\ \ y_{chastn}=(Ax^2+Bx+C)\cdot x^{r}\cdot e^{4x}\\\\ y_{chastn}=(Ax^2+Bx+C)\cdot x\cdot e^{4x}=(Ax^3+Bx^2+Cx)\cdot e^{4x}\\\\ y'_{chastn}=(3Ax^2+2Bx+C)\cdot e^{4x}+(Ax^3+Bx^2+Cx)\cdot 4e^{4x}$

$y''_{chastn}=(6Ax+2B)\cdot e^{4x}+(3Ax^2+Bx+C)\cdot 4e^{4x}+\\{}\ \ \ \ \ \ \ \ \ \ \ +(3Ax^2+2Bx+C)\cdot 4e^{4x}+(Ax^3+Bx^2+Cx)\cdot 16e^{4x}\\-------------------------------$

$y''-9y'+20y=-(3Ax^2+2Bx+C)e^{4x}+(6Ax+2B)e^{4x}=x^2e^{4x}\\\\-3Ax^2-2Bx-C+6Ax+2B=x^2\\\\x^2\ |\ -3A=1\ \ ,\qquad A=-\dfrac{1}{3}\\x\ \ |\ -2B+6A=0\ \ ,\ \ B=3A\ \ ,\ \ B=-3\cdot \dfrac{1}{3}=-1\\x^0\ |\ -C+2B=0\ \ ,\ \ C=2B=-2\\\\\\y_{chastn}=\Big(-\dfrac{1}{3}\, x^2-x^2-2x\Big)\cdot e^{4x}\\\\\\c)\ \ y_{o.n.}=C_1e^{4x}+C_2e^{5x}-\Big(\dfrac{1}{3}\, x^2+x^2+2x\Big)\cdot e^{4x}$

$2)\ \ y''-y'=cos2x+sin2x\\\\a)\ \ k^2-k=0\ \ ,\ \ k\, (k-1)=0\ \ ,\ \ k_1=0\ ,\ k_2=1\\\\y_{oo}=C_1+C_2e^{x}\\\\b)\ \ f(x)=e^{0\cdot x}(cos2x+sin2x)\ \ ,\ \ \alpha =0\ ,\ \beta =2\ ,\ \alpha +\beta i=2i\ne k_{1,2}\ \ \to \ \ r=0\\\\y_{chastn}=Acos2x+Bsin2x\\\\y'_{chastn}=-2Asin2x+2Bcos2x\\\\y''_{chastn}=-4Acos2x-4Bsin2x\\----------------------------------\\y''-y'=-4Acos2x-4Bsin2x-(-2Asin2x+2Bcos2x)=cos2x+sin2x\\\\cos2x\ |\ -4A-2B=1\ \ ,\qquad \\sin2x\ |\ -4B+2A=1\ \ ,$

$\left\{\begin{array}{l}4A+2B=-1\\2A-4B=1\ |\cdot (-2)\end{array}\right\ \oplus \ \left\{\begin{array}{l}-6B=-3\\2A=1+4B\end{array}\right\ \ \left\{\begin{array}{l}B=0,5\\A=\frac{1}{2}(1+2)=1,5\end{array}\right\\\\\\y_{chastn}=1,5\cdot cos2x+0,5\cdot sin2x\\\\c)\ \ y_{o.n.}=C_1+C_2e^{x}+1,5\cdot cos2x+0,5\cdot sin2x$