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 $y = 1$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ when $x = 0$

(5)

解答

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这是二阶常系数非齐次微分方程。先求 complementary function，再设 particular integral。右边是二次多项式 $2x^2+x$ ，所以 particular integral 也设成二次多项式。

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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*}

giving

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

Hence 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=\lambda x^2+\mu x+\nu. \end{align*}

Then

\begin{align*} y_p'=&\,2\lambda x+\mu,\\[2mm] y_p''=&\,2\lambda. \end{align*}

Substitute into the differential equation:

\begin{align*} 2\lambda -6(2\lambda x+\mu) +8(\lambda x^2+\mu x+\nu) =&\,2x^2+x. \end{align*}

Compare coefficients:

\begin{align*} x^2:\quad 8\lambda=&\,2,\\[2mm] x:\quad -12\lambda+8\mu=&\,1,\\[2mm] \text{constant}:\quad 2\lambda-6\mu+8\nu=&\,0. \end{align*}

Solving,

\begin{align*} \lambda=\frac14,\qquad \mu=\frac12,\qquad \nu=\frac5{16}. \end{align*}

Therefore,

\begin{align*} y =&\,A\mathrm{e}^{2x}+B\mathrm{e}^{4x} +\frac14x^2+\frac12x+\frac5{16}. \end{align*}

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把 $x=0$ 、 $y=1$ 代入通解得到一个关于 $A,B$ 的方程。再对通解求导，用 $x=0$ 、 $y'=0$ 得到第二个方程。

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

\begin{align*} 1=A+B+\frac5{16}. \end{align*}

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

Differentiate the general solution:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,2A\mathrm{e}^{2x}+4B\mathrm{e}^{4x} +\frac12x+\frac12. \end{align*}

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

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

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

Solve the simultaneous equations:

\begin{align*} A+B=&\,\frac{11}{16},\\[2mm] A+2B=&\,-\frac14. \end{align*}

Subtracting gives

\begin{align*} B=-\frac{15}{16}. \end{align*}

Then

\begin{align*} A=\frac{11}{16}+\frac{15}{16}=\frac{13}{8}. \end{align*}

Therefore the particular solution is

\begin{align*} y =&\,\frac{13}{8}\mathrm{e}^{2x} -\frac{15}{16}\mathrm{e}^{4x}\\[4mm] &\,\hspace{2pt}+\frac14x^2+\frac12x+\frac5{16}. \end{align*}