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# 如何用降阶法求二阶常系数线性微分方程的解

$y^{\prime\prime}-5y’+6y=xe^x$

\begin{align*}& y ^{\prime\prime} -2y’-3y’+6y=xe^x \\ \Longrightarrow& (y ^{\prime\prime} -2y’)-3(y’-2y)=xe^x \\ \Longrightarrow & (y’-2y)’-3(y’-2y)=xe^x\\ \end{align*}

$z’-3z=xe^x$

\begin{align*}z&=e^{3x}\left(\int e^{-3x}xe^xdx+ C_1\right)\\ &= C_1e^{3x}-\frac{1}{2}xe^{-x}-\frac{1}{4}e^{-x}\end{align*}

$y’-2y= C_1e^{3x}-\frac{1}{2}xe^{-x}-\frac{1}{4}e^{-x}$

\begin{align*}y&=e^{2x}\left(\int e^{-2x}( C_1e^{3x}-\frac{1}{2}xe^{-x}-\frac{1}{4}e^{-x} )dx+C_2\right)\\ &=C_1e^{3x}+C_2e^{2x}+\frac{1}{6}xe^{-3x}+\frac{1}{18}e^{-3x}+\frac{1}{12}e^{-3x}\\ &= C_1e^{3x}+C_2e^{2x}+\frac{1}{6}xe^{-3x}+ \frac{5}{36}e^{-3x}\end{align*}

$y^{\prime\prime}+py’+qy=f(x)$

$\qquad y^{\prime\prime}-(\lambda_1+\lambda_2)y’+\lambda_1\lambda_2y=f(x)$

$\Longrightarrow(y’-\lambda_2y)’-\lambda_1( y’-\lambda_2y )=f(x)$

$y^{\prime\prime}-4y’+4y=e^{2x}\sin x$

\begin{align*}z&=e^{2x}\left(\int e^{-2x} e^{2x}\sin x dx+C_1\right)\\ &=C_1e^{2x}-e^{2x}\cos x\end{align*}

$y’-2y= C_1e^{2x}-e^{2x}\cos x$

\begin{align*}y&=e^{2x}\left(\int e^{-2x}( C_1e^{2x}-e^{2x}\cos x )dx+C_2\right)\\ &=C_1xe^{2x}+C_2e^{2x}-e^{2x}\sin x\end{align*}

$y^{\prime\prime}+4y=e^x$

$(y’-2iy)’+2i(y’-2iy)=e^x$

\begin{align*}z&=e^{-2ix}\left(\int e^{2ix}e^xdx+C_1 \right) \\ &=C_1 e^{-2ix} +\frac{1 }{1+2i} e^x \end{align*}

$y-2iy= C_1 e^{-2ix} +\frac{1 }{1+2i} e^x$

\begin{align*}y&= e^{2ix}\left(\int e^{-2ix}\left( C_1 e^{-2ix} +\frac{1 }{1+2i} e^x dx\right)+C_2 \right)\\ &=C_1e^{-2ix}+C_2e^{2ix}+\frac{1}{5}e^x \end{align*}

\begin{align*}y&=C_1(\cos 2x-i\sin 2x)+C_2(\cos 2x+i\sin 2x)+ \frac{1}{5}e^x \\ &=(C_1+C_2)\cos 2x+i(C_2-C_1)\sin2x+ \frac{1}{5}e^x \\&=\tilde{C_1}\cos 2x+\tilde{C_2}\sin 2x+ \frac{1}{5}e^x \end{align*}

$y=C_1\cos2x+C_2\sin2x+ \frac{1}{5}e^x$