東京大学 新領域創成科学研究科 人間環境学専攻 2024年8月実施 微分方程

标签：

• Tokyo-University
• Second-Order-Linear-Ordinary-Differential-Equation

Author

思齐塾, 祭音Myyura

Description

以下の微分方程式 1), 2), 3), 4) について一般解をそれぞれ求めよ。

1. $\frac{d^3y}{dx^3} - 9\frac{d^2y}{dx^2} + 27\frac{dy}{dx} - 27y = 0$

2. $\frac{d^2y}{dx^2} + y = \sin 3x$

3. $\frac{d^3y}{dx^3} - 3\frac{d^2y}{dx^2} = 0$

4. $(x-y)dx + (2x^2y + x)dy + 4x^2zdz = 0$

题目描述

分别求下列四个微分方程的通解：

1. 求解：

   $$\frac{d^3y}{dx^3} -9\frac{d^2y}{dx^2} +27\frac{dy}{dx} -27y=0.$$

2. 求解：

   $$\frac{d^2y}{dx^2}+y=\sin 3x.$$

3. 求解：

   $$\frac{d^3y}{dx^3} -3\frac{d^2y}{dx^2}=0.$$

4. 求解：

   $$(x-y)\,dx+(2x^2y+x)\,dy+4x^2z\,dz=0.$$

Kai

1. The characteristic equation is $r^3 - 9r^2 + 27r - 27 = 0$ , which can be factored as $(r-3)^3 = 0$ . Thus, $r = 3$ is a triple root. The general solution is $y(x) = c_1e^{3x} + c_2xe^{3x} + c_3x^2e^{3x}$ .

2. The homogeneous equation is $\frac{d^2y}{dx^2} + y = 0$ . The characteristic equation is $r^2 + 1 = 0$ , so $r = \pm i$ . The homogeneous solution is $y_h(x) = c_1\cos x + c_2\sin x$ . For the particular solution, let $y_p(x) = A\cos 3x + B\sin 3x$ . Then $y_p'(x) = -3A\sin 3x + 3B\cos 3x$ and $y_p''(x) = -9A\cos 3x - 9B\sin 3x$ . Substituting into the differential equation gives $-9A\cos 3x - 9B\sin 3x + A\cos 3x + B\sin 3x = \sin 3x$ . Thus, $-8A = 0$ and $-8B = 1$ , so $A = 0$ and $B = -\frac{1}{8}$ . Therefore, $y_p(x) = -\frac{1}{8}\sin 3x$ . The general solution is $y(x) = c_1\cos x + c_2\sin x - \frac{1}{8}\sin 3x$ .

3. The characteristic equation is $r^3 - 3r^2 = 0$ , so $r^2(r-3) = 0$ . The roots are $r = 0$ (double root) and $r = 3$ . The general solution is $y(x) = c_1 + c_2x + c_3e^{3x}$ .

4. $x\ne0$ の領域で積分因子 $\mu=x^{-2}$ を掛けると

$$\left(\frac1x-\frac{y}{x^2}\right)dx +\left(2y+\frac1x\right)dy+4z\,dz=0.$$

この微分形式は完全であり、

$$\frac{\partial}{\partial x} \left(\log|x|+\frac{y}{x}+y^2+2z^2\right) =\frac1x-\frac{y}{x^2},$$

$$\frac{\partial}{\partial y} \left(\log|x|+\frac{y}{x}+y^2+2z^2\right) =\frac1x+2y, \qquad \frac{\partial}{\partial z} \left(\log|x|+\frac{y}{x}+y^2+2z^2\right) =4z.$$

したがって一般積分は

$$\boxed{\log|x|+\frac{y}{x}+y^2+2z^2=C}\qquad(x\ne0).$$

また $x=0$ は、もとの微分形式の引き戻しが恒等的に零となる特異積分面である。