题目

Problem

(a) Find the general solution of the differential equation

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} -6\frac{\mathrm{d}y}{\mathrm{d}x} +8y = 2x^2+x \end{align*}

(8)

(b) Find the particular solution of this differential equation for which

\begin{align*} y=1 \quad\text{and}\quad \frac{\mathrm{d}y}{\mathrm{d}x}=0 \quad\text{when }x=0 \end{align*}

(5)

解答

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这是常系数非齐次二阶微分方程。先解辅助方程得到互补函数；右边是二次多项式，所以特解设为 $\alpha x^2+\beta x+\gamma$ 。

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The auxiliary equation is

\begin{align*} m^2-6m+8=0 \end{align*}

\begin{align*} (m-2)(m-4)=0 \end{align*}

Thus

\begin{align*} m=2,\quad 4 \end{align*}

The complementary function is

\begin{align*} y_c=A\mathrm{e}^{2x}+B\mathrm{e}^{4x} \end{align*}

For a particular integral, let

\begin{align*} y_p=\alpha x^2+\beta x+\gamma \end{align*}

Then

\begin{align*} y_p'=2\alpha x+\beta, \qquad y_p''=2\alpha \end{align*}

Substitute into the differential equation:

\begin{align*} 2\alpha-6(2\alpha x+\beta) +8(\alpha x^2+\beta x+\gamma) = 2x^2+x \end{align*}

Compare coefficients:

\begin{align*} 8\alpha=&\,2\\[2mm] -12\alpha+8\beta=&\,1\\[2mm] 2\alpha-6\beta+8\gamma=&\,0 \end{align*}

\begin{align*} \alpha=\frac14, \qquad \beta=\frac12, \qquad \gamma=\frac{5}{16} \end{align*}

Therefore the general solution is

\begin{align*} y = A\mathrm{e}^{2x}+B\mathrm{e}^{4x} +\frac14x^2+\frac12x+\frac{5}{16} \end{align*}

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把 $x=0$ 、 $y=1$ 代入通解得到第一个方程；再求导并用 $y'(0)=0$ 得到第二个方程，联立求 $A,B$ 。

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From part (a),

\begin{align*} y = A\mathrm{e}^{2x}+B\mathrm{e}^{4x} +\frac14x^2+\frac12x+\frac{5}{16} \end{align*}

Using $y=1$ when $x=0$ ,

\begin{align*} 1=A+B+\frac{5}{16} \end{align*}

\begin{align*} A+B=\frac{11}{16} \end{align*}

Differentiate:

\begin{align*} y' = 2A\mathrm{e}^{2x}+4B\mathrm{e}^{4x} +\frac12x+\frac12 \end{align*}

Using $y'=0$ when $x=0$ ,

\begin{align*} 0 = 2A+4B+\frac12 \end{align*}

\begin{align*} 2A+4B=-\frac12 \end{align*}

Solving

\begin{align*} A+B=\frac{11}{16}, \qquad 2A+4B=-\frac12, \end{align*}

gives

\begin{align*} A=\frac{13}{8}, \qquad B=-\frac{15}{16} \end{align*}

Therefore

\begin{align*} y = \frac{13}{8}\mathrm{e}^{2x} -\frac{15}{16}\mathrm{e}^{4x} +\frac14x^2+\frac12x+\frac{5}{16} \end{align*}