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東京大学 工学系研究科 2016年8月実施 数学 第1問

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I、以下の定積分を求めよ。

I=∫24dx(x−2)(4−x)\begin{align} I = \int_2^4 \frac{\text{d}x}{\sqrt{(x-2)(4-x)}} \tag{1} \end{align}

II、以下の微分方程式の一般解と特異解を求めよ。

y=xdydx+dydx+(dydx)2\begin{align} y = x \frac{\text{d}y}{\text{d}x} + \frac{\text{d}y}{\text{d}x} + \bigg(\frac{\text{d}y}{\text{d}x} \bigg)^2 \tag{2} \end{align}

III、以下の微分方程式の一般解を求めよ。

x2d2ydx2−xdydx−8y=x2\begin{align} x^2 \frac{\text{d}^2y}{\text{d}x^2} - x \frac{\text{d}y}{\text{d}x} - 8y = x^2 \tag{3} \end{align}

Kai​

(I)​

I=∫24dx(x−2)(4−2)=∫24dx1−(x−3)2=∫−π/2π/2cos⁡θdθ1−sin⁡2θ(置換:x−3=sin⁡θ)=∫−π/2π/2dθ=π2−(−π2)=π\begin{aligned} I &= \int_2^4 \frac{\text{d}x}{\sqrt{(x-2)(4-2)}} \\ &=\int_2^4 \frac{\text{d}x}{\sqrt{1-(x-3)^2}} \\ &=\int_{-\pi /2}^{\pi /2} \frac{\cos\theta \text{d}\theta}{\sqrt{1-\sin^2 \theta}} \qquad (\text{置換:} \quad x - 3 = \sin \theta) \\ &=\int_{-\pi /2}^{\pi /2} \text{d}\theta = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi \end{aligned}

(II)​

y=xdydx+dydx+(dydx)2y=xy′+y′+(y′)2\begin{aligned} y = x \frac{\text{d}y}{\text{d}x} + \frac{\text{d}y}{\text{d}x} + \bigg(\frac{\text{d}y}{\text{d}x} \bigg)^2 \\ y = xy' + y' + (y')^2 \\ \end{aligned}

xで微分してx\text{で微分して}

y′=y′+xy′′+y′′+2y′y′′y′′(x+1+2y′)=0\begin{aligned} y' = y' + xy'' + y'' + 2y'y''\\ y''(x + 1 + 2y') = 0 \end{aligned}

(i)​

y′′=0y'' = 0 のとき,

y=ax+by = ax + b

となり,　式(22)に代入すると,

ax+b=a(x+1)+a2b=a2+a\begin{aligned} ax + b &= a(x + 1) + a^2 \\ b &= a^2 + a \end{aligned}

である,　よって一般解 y=ax+a2+ay = ax + a^2 + a を得る。

(ii)​

x+1+2y′=0x + 1 + 2y' = 0 のとき,

y′=−12(x+1)y=−14x2−12x+C\begin{aligned} y' &= -\frac{1}{2}(x + 1) \\ y &= -\frac{1}{4}x^2 - \frac{1}{2}x + C \\ \end{aligned}

となり,　式(22)に代入すると,

−14x2−12x+C=(x+1)(−12x−12)+(−12x−12)2C=−14\begin{aligned} - \frac{1}{4}x^2 - \frac{1}{2}x + C &= (x + 1)(-\frac{1}{2}x - \frac{1}{2}) + (-\frac{1}{2}x - \frac{1}{2})^2 \\ C &= -\frac{1}{4} \end{aligned}

である,　よって特異解 y=−14x2−12x−14y = -\frac{1}{4}x^2 - \frac{1}{2}x - \frac{1}{4} を得る。

(III)​

x2d2ydx2−xdydx−8y=x2\begin{aligned} x^2 \frac{\text{d}^2y}{\text{d}x^2} - x \frac{\text{d}y}{\text{d}x} - 8y = x^2 \end{aligned}

x=etx = e^{t} とおくと,

dxdt=et=x,dtdx=1xxdydx=xdydtdtdx=dydtd2ydx2=ddx(1xdydt)=−1x2dydt+1xddtdtdxdydt=−1x2dytextdt+1x2d2ydt2∴x2d2ydx2=d2ydt2−dydt\begin{aligned} \frac{\text{d}x}{\text{d}t} &= e^{t} = x , \qquad \frac{\text{d}t}{\text{d}x} = \frac{1}{x} \\ x \frac{\text{d}y}{\text{d}x}&= x \frac{\text{d}y}{\text{d}t} \frac{\text{d}t}{\text{d}x} = \frac{\text{d}y}{\text{d}t} \\ \frac{\text{d}^2y}{\text{d}x^2} &= \frac{\text{d}}{\text{d}x} (\frac{1}{x} \frac{\text{d}y}{\text{d}t}) \\ &= - \frac{1}{x^2} \frac{\text{d}y}{\text{d}t} + \frac{1}{x} \frac{\text{d}}{\text{d}t} \frac{\text{d}t}{\text{d}x} \frac{\text{d}y}{\text{d}t} \\ &= -\frac{1}{x^2} \frac{\text{d}y}{text{d}t} + \frac{1}{x^2} \frac{\text{d}^2y}{\text{d}t^2} \\ &\therefore x^2\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}^2y}{\text{d}t^2} - \frac{\text{d}y}{\text{d}t} \\ \end{aligned}

であるから,式(33)は,

d2ydt2−dydt−dydt−8y=e2td2ydt2−2dydt−8y=e2t\begin{align} \frac{\text{d}^2y}{\text{d}t^2} - \frac{\text{d}y}{\text{d}t} - \frac{\text{d}y}{\text{d}t} - 8y &= e^{2t} \nonumber \\ \frac{\text{d}^2y}{\text{d}t^2} - 2\frac{\text{d}y}{\text{d}t} - 8y &= e^{2t} \tag{4} \end{align}

となる。特性方程式 λ2−2λ−8=0\lambda^2 - 2\lambda - 8 = 0 の解は,

(λ−4)(λ+2)=0∴λ=−2,4(\lambda - 4)(\lambda + 2) = 0 \\ \therefore \lambda = -2 , \quad 4

だから,　斉次の一般解は y=C1e−2t+C2e4ty = C_{1}e^{-2t} + C_{2}e^{4t}。

一方,　特解を y=Ae2ty = Ae^{2t} と予想して式(44)に代入すると,

4Ae2t−4Ae2t−8Ae2t=e2t∴A=−18\begin{aligned} 4Ae^{2t} &- 4Ae^{2t} - 8Ae^{2t} = e^{2t} \\ &\therefore A = -\frac{1}{8} \end{aligned}

となり,　特解 y=−18e2ty = -\frac{1}{8}e^{2t} を得る。よって求める一般解は,

y=C1e−2t+C2e4t−18e2t=C1x−2+C2x4−18x\begin{aligned} y &= C_{1}e^{-2t} + C_{2}e^{4t} - \frac{1}{8}e^{2t} \\ &= C_{1}x^{-2} + C_{2}x^4 - \frac{1}{8}x \\ \end{aligned}
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