题目

Problem

(a) Determine the general solution of the differential equation

\begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -4\frac{\mathrm{d}y}{\mathrm{d}x} +37y =&\, 6\mathrm{e}^{5x} \end{align*}

(6)

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

(b) determine the particular solution for this differential equation.

(5)

题目中文翻译

(a) 求微分方程

\begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -4\frac{\mathrm{d}y}{\mathrm{d}x}+37y =&\,6\mathrm{e}^{5x} \end{align*}

的通解。

(6)

已知 $x=0$ 时， $y=0$ 且 $\frac{\mathrm{d}y}{\mathrm{d}x}=0$ 。

(b) 求这个微分方程的特解。

(5)

解答

(a)

解法一

思路

展开

这是二阶常系数非齐次微分方程。先用辅助方程求 complementary function，再因为右边是 $6\mathrm{e}^{5x}$ ，设 particular integral 为 $k\mathrm{e}^{5x}$ 。

答题过程

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The auxiliary equation is

\begin{align*} 4m^2-4m+37=0 \end{align*}

Solve:

\begin{align*} m =&\, \frac{4\pm\sqrt{(-4)^2-4(4)(37)}}{8}\\[3mm] =&\, \frac{4\pm\sqrt{-576}}8\\[3mm] =&\, \frac{4\pm24\mathrm{i}}8\\[3mm] =&\, \frac12\pm3\mathrm{i} \end{align*}

So the complementary function is

\begin{align*} y_{\mathrm{c}} =&\, \mathrm{e}^{x/2}\bigg(A\cos3x+B\sin3x\bigg) \end{align*}

For the particular integral, try

\begin{align*} y_{\mathrm{p}}=k\mathrm{e}^{5x} \end{align*}

Then

\begin{align*} y_{\mathrm{p}}' =&\, 5k\mathrm{e}^{5x},\\ y_{\mathrm{p}}'' =&\, 25k\mathrm{e}^{5x} \end{align*}

Substitute into the differential equation:

\begin{align*} 4(25k\mathrm{e}^{5x}) -4(5k\mathrm{e}^{5x}) +37(k\mathrm{e}^{5x}) =&\, 6\mathrm{e}^{5x}\\[3mm] 117k\mathrm{e}^{5x} =&\, 6\mathrm{e}^{5x}\\[3mm] k =&\, \frac{6}{117} =\frac2{39} \end{align*}

Therefore the general solution is

\begin{align*} \boxed{ y= \mathrm{e}^{x/2}\bigg(A\cos3x+B\sin3x\bigg) +\frac2{39}\mathrm{e}^{5x} } \end{align*}

(b)

解法一

思路

展开

把 $x=0,y=0$ 代入通解可先求 $A$ 。然后对通解求导，再用 $x=0,y'=0$ 求 $B$ 。

答题过程

展开

From part (a),

\begin{align*} y= \mathrm{e}^{x/2}\bigg(A\cos3x+B\sin3x\bigg) +\frac2{39}\mathrm{e}^{5x} \end{align*}

Use $y=0$ when $x=0$ :

\begin{align*} 0 =&\, A+\frac2{39}\\[2mm] A =&\, -\frac2{39} \end{align*}

Differentiate the general solution:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\, \frac12\mathrm{e}^{x/2} \bigg(A\cos3x+B\sin3x\bigg)\\[2mm] &\,\hspace{2pt}+\mathrm{e}^{x/2} \bigg(-3A\sin3x+3B\cos3x\bigg)\\[2mm] &\,\hspace{4pt}+\frac{10}{39}\mathrm{e}^{5x} \end{align*}

Use $\frac{\mathrm{d}y}{\mathrm{d}x}=0$ when $x=0$ :

\begin{align*} 0 =&\, \frac12A+3B+\frac{10}{39}\\[2mm] =&\, \frac12\bigg(-\frac2{39}\bigg)+3B+\frac{10}{39}\\[2mm] =&\, -\frac1{39}+3B+\frac{10}{39}\\[2mm] 3B =&\, -\frac9{39}\\[2mm] B =&\, -\frac1{13} \end{align*}

Hence the particular solution is

\begin{align*} \boxed{ y= \frac2{39}\mathrm{e}^{5x} +\mathrm{e}^{x/2} \bigg( -\frac2{39}\cos3x-\frac1{13}\sin3x \bigg) } \end{align*}

FP2 2025 Jan Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程