高専の数学解説

第10章 10.11

(1)

x2y′′+xy′+y=xx^2 y'' + xy' + y = x

x=etx = e^t とおくと y′′+y=ety'' + y = e^t

特性方程式は s2+1=0s^2 + 1 = 0 ∴s=±i\therefore s = \pm i

微分演算子 D を使って (D2+1)y1=et(D^2 + 1)y_1 = e^t y1=1D2+1et=et2y_1 = \frac{1}{D^2+1}e^t = \frac{e^t}{2} これから

y=Asin⁡t+Bcos⁡t+et2y = A\sin t + B\cos t + \frac{e^t}{2} 元に戻して

y=Asin⁡(log⁡x)+Bcos⁡(log⁡x)+x2y = A\sin(\log x) + B\cos(\log x) + \frac{x}{2} （A,B は任意定数）

(2)

x2y′′−2xy′+2y=−2x+2x^2 y'' - 2xy' + 2y = -2x + 2 y′′−2xy′+2x2y=−2x+2x2y'' - \frac{2}{x}y' + \frac{2}{x^2}y = -\frac{2}{x} + \frac{2}{x^2}

P=−2xP = -\frac{2}{x}、Q=2x2Q = \frac{2}{x^2}、R=−2x+2x2R = -\frac{2}{x} + \frac{2}{x^2} とおくと y=exp⁡(−12∫−2xdx)v=xvy = \exp\left(-\frac{1}{2}\int -\frac{2}{x}dx\right) v = xv

I=2x2−12(2x2)−14(4x2)=0I = \frac{2}{x^2} - \frac{1}{2}\left(\frac{2}{x^2}\right) - \frac{1}{4}\left(\frac{4}{x^2}\right) = 0 J=−2x2+2x3J = -\frac{2}{x^2} + \frac{2}{x^3} これから

v′′=−2x2+2x3v'' = -\frac{2}{x^2} + \frac{2}{x^3} v′=2x−1x2+C1v' = \frac{2}{x} - \frac{1}{x^2} + C_1 v=2log⁡x+1x+C1x+C2v = 2\log x + \frac{1}{x} + C_1 x + C_2

元に戻して C1=AC_1 = A、C2=BC_2 = B とおくと

y=2xlog⁡x+1+Ax2+Bxy = 2x\log x + 1 + Ax^2 + Bx （A,B は任意定数）

(3)

(1+x2)y′′−2xy′+2y=1−x2x(1 + x^2)y'' - 2xy' + 2y = \frac{1-x^2}{x}

(1+x2)y′′−2xy′+2y=0(1 + x^2)y'' - 2xy' + 2y = 0 の式として考えると

m(m−1)(1+x2)−2mx2+2x2=0m(m-1)(1+x^2) - 2mx^2 + 2x^2 = 0 (m−1){m(1−x2)−2x2}=0(m-1)\{m(1-x^2) - 2x^2\} = 0 m=1m = 1

これから y=xy = x を解に持つ

y=xvy = xv とおくと y′=v+xv′y' = v + xv' y′′=2v′+xv′′y'' = 2v' + xv'' これから与式は

(1+x2)(2v′+xv′′)−2x(v+xv′)+2xv=1−x2x(1+x^2)(2v'+xv'') - 2x(v+xv') + 2xv = \frac{1-x^2}{x} x(1−x2)v′′+2v′=1−x2xx(1-x^2)v'' + 2v' = \frac{1-x^2}{x}

v′′+2x(1+x2)v′=1−x2x2(1+x2)v'' + \frac{2}{x(1+x^2)} v' = \frac{1-x^2}{x^2(1+x^2)}

v′=Pv' = P とおくと

P′+2x(1+x2)P=1−x2x2(1+x2)P' + \frac{2}{x(1+x^2)} P = \frac{1-x^2}{x^2(1+x^2)}

α=2x(1+x2)、β=1−x2x2(1+x2)\alpha = \frac{2}{x(1+x^2)}、\beta = \frac{1-x^2}{x^2(1+x^2)} とおくと

∫2x(1+x2)dx=∫(2x−2x1+x2)dx=2log⁡x−log⁡(1+x2)\int \frac{2}{x(1+x^2)} dx = \int \left(\frac{2}{x} - \frac{2x}{1+x^2}\right) dx = 2\log x - \log(1+x^2) これから

P=1+x2x2{∫x21+x2⋅1−x2x2(1+x2)dx+C1}=1+x2x2{∫1−x2(1+x2)2dx+C1}P = \frac{1+x^2}{x^2} \left\{\int \frac{x^2}{1+x^2} \cdot \frac{1-x^2}{x^2(1+x^2)} dx + C_1\right\} = \frac{1+x^2}{x^2} \left\{\int \frac{1-x^2}{(1+x^2)^2} dx + C_1\right\}

=1+x2x2{∫1(x2+1)dx+∫−x2(x2+1)2dx+C1}= \frac{1+x^2}{x^2} \left\{\int \frac{1}{(x^2+1)} dx + \int \frac{-x^2}{(x^2+1)^2} dx + C_1\right\}

=1+x2x2{12(xx2+1+tan⁡−1x+xx2+1−tan⁡−1x)+C1}= \frac{1+x^2}{x^2} \left\{\frac{1}{2}\left(\frac{x}{x^2+1} + \tan^{-1} x + \frac{x}{x^2+1} - \tan^{-1} x\right) + C_1\right\}

=1+x2x2(xx2+1+C1)=1x+1+x2x2C1= \frac{1+x^2}{x^2}\left(\frac{x}{x^2+1} + C_1\right) = \frac{1}{x} + \frac{1+x^2}{x^2}C_1 元に戻して

v′=1x+1+x2x2C1v' = \frac{1}{x} + \frac{1+x^2}{x^2} C_1 v=log⁡x−(1−xx)C1+C2v = \log x - \left(\frac{1-x}{x}\right) C_1 + C_2 元に戻して

y=xlog⁡x−(1−x2)C1+C2xy = x\log x - (1-x^2)C_1 + C_2 x C1=B、C2=AC_1 = B、C_2 = A とおくと

y=xlog⁡x−B(1−x2)+Axy = x\log x - B(1-x^2) + Ax （A,B は任意定数）

(4)

x4y′′+2x3y′+y=x+1xx^4 y'' + 2x^3 y' + y = \frac{x+1}{x}

t=1xt = \frac{1}{x} とおくと t′=−1x2t' = -\frac{1}{x^2} dydx=dydtdtdx=−t2dydt\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = -t^2\frac{dy}{dt}

d2ydx2=ddt(−t2dydt)dxdt=(−2tdydt−t2d2ydt2)(−t2)=t4d2ydt2+2t3dydt\frac{d^2y}{dx^2} = \frac{d}{dt}\left(-t^2\frac{dy}{dt}\right)\frac{dx}{dt} = \left(-2t\frac{dy}{dt} - t^2\frac{d^2y}{dt^2}\right)(-t^2) = t^4\frac{d^2y}{dt^2} + 2t^3\frac{dy}{dt} これから与式は

x4(t4d2ydt2+2t3dydt)+2x3(−t2dydt)+y=x+1xx^4\left(t^4\frac{d^2y}{dt^2} + 2t^3\frac{dy}{dt}\right) + 2x^3\left(-t^2\frac{dy}{dt}\right) + y = \frac{x+1}{x}

1t4(t4d2ydt2+2t3dydt)+2t3(−t2dydt)+y=t(1t+1)\frac{1}{t^4}\left(t^4\frac{d^2y}{dt^2}+2t^3\frac{dy}{dt}\right)+\frac{2}{t^3}\left(-t^2\frac{dy}{dt}\right)+y=t\left(\frac{1}{t}+1\right)

d2ydt2+2tdydt+2tdydt+y=1+td2ydt2+y=1+t\frac{d^2y}{dt^2}+\frac{2}{t}\frac{dy}{dt}+\frac{2}{t}\frac{dy}{dt}+y=1+t \quad \frac{d^2y}{dt^2}+y=1+t

特性方程式は s2+1=0s^2+1=0 s=±is=\pm i

微分演算子Dを使って (D2+1)y1=t+1(D^2+1)y_1=t+1 y1=t+1D2+1=t+1y_1=\frac{t+1}{D^2+1}=t+1 これから

y=Asin⁡t+Bcos⁡t+t+1y=A\sin t+B\cos t+t+1 元に戻して

y=Asin⁡1x+Bcos⁡1x+1x+1y=A\sin\frac{1}{x}+B\cos\frac{1}{x}+\frac{1}{x}+1 （A,BA,Bは任意定数）

(5)

xy′′−(2x+1)y′+(x+1)y=(x2+x−1)e2xxy''-(2x+1)y'+(x+1)y=(x^2+x-1)e^{2x}

P=x,Q=−(2x+1),R=x+1P=x, Q=-(2x+1), R=x+1として

m2x+m(2x+1)+x+1=0m^2x+m(2x+1)+x+1=0 {mx−(x+1)}(m−1)=0\{mx-(x+1)\}(m-1)=0 m=1m=1

よってy=exyy=e^xyとおくと y′=exy+exy′y'=e^xy+e^xy' y′′=exy+2exy′+exy′′y''=e^xy+2e^xy'+e^xy'' これより与式は

x(exy+2exy′+exy′′)−(2x+1)(exy+exy′)+(x+1)exy=(x2+x−1)e2xx(e^xy+2e^xy'+e^xy'')-(2x+1)(e^xy+e^xy')+(x+1)e^xy=(x^2+x-1)e^{2x}

xy′′−y′=(x2+x−1)exxy''-y'=(x^2+x-1)e^x y′=Puy'=Puとおくと

xP′−P=(x2+x−1)exxP'-P=(x^2+x-1)e^x P−1xP=x2+x−1xexP-\frac{1}{x}P=\frac{x^2+x-1}{x}e^x

α=−1x,β=x2+x−1xex\alpha=-\frac{1}{x}, \beta=\frac{x^2+x-1}{x}e^xとおくと ∫αdx=−log⁡x\int\alpha dx=-\log x これから

P=x{∫1x⋅x2+x−1xexdx+C1}=x{∫[1+1x−1x2]exdx+C1}=x{∫exdx+∫(1x−1x2)exdx+C1}P=x\left\{\int\frac{1}{x}\cdot\frac{x^2+x-1}{x}e^xdx+C_1\right\}=x\left\{\int\left[1+\frac{1}{x}-\frac{1}{x^2}\right]e^xdx+C_1\right\}=x\left\{\int e^xdx+\int\left(\frac{1}{x}-\frac{1}{x^2}\right)e^xdx+C_1\right\}

ここで∫(1x−1x2)exdx=∫1xexdx−∫1x2exdx=1xex+∫1x2exdx+1xex−∫1xexdx\int\left(\frac{1}{x}-\frac{1}{x^2}\right)e^xdx=\int\frac{1}{x}e^xdx-\int\frac{1}{x^2}e^xdx=\frac{1}{x}e^x+\int\frac{1}{x^2}e^xdx+\frac{1}{x}e^x-\int\frac{1}{x}e^xdx

=2xex−∫(1x−1x2)exdx=\frac{2}{x}e^x-\int\left(\frac{1}{x}-\frac{1}{x^2}\right)e^xdx

∫(1x−1x2)exdx=1xexこれから\int\left(\frac{1}{x}-\frac{1}{x^2}\right)e^xdx=\frac{1}{x}e^x \quad \text{これから}

P=x{ex+1xex+C1}P = x\{e^x + \frac{1}{x}e^x + C_1\} 元に戻して

y′=xex+ex+xC1y' = xe^x + e^x + xC_1 y=xex−∫exdx+ex+12C1x2+C2=xex+12C1x2+C2y = xe^x - \int e^x dx + e^x + \frac{1}{2}C_1 x^2 + C_2 = xe^x + \frac{1}{2}C_1 x^2 + C_2 元に戻して

y=xe2x+ex(12C1x2+C2)y = xe^{2x} + e^x(\frac{1}{2}C_1 x^2 + C_2) 12C1=A,C2=B\frac{1}{2}C_1 = A, C_2 = B とおくと

y=xe2x+ex(Ax2+B)y = xe^{2x} + e^x(Ax^2 + B) (A,BA,B は任意定数)

(6)

(x+1)y′′−(3x+4)y′+3y=(3x+2)e3x(x+1)y'' - (3x+4)y' + 3y = (3x+2)e^{3x}

P=x+1P = x+1、Q=−(3x+4)Q = -(3x+4)、R=3R = 3 とおくと

m2(x+1)−m(3x+4)+3=0m^2(x+1) - m(3x+4) + 3 = 0 {m(x+1)−1}(m−3)=0\{m(x+1) - 1\}(m - 3) = 0 m=3m = 3

これから y=e3xvy = e^{3x}v とおくと y′=3e3xv+e3xv′y' = 3e^{3x}v + e^{3x}v' y′′=9e3xv+6e3xv′+e3xv′′y'' = 9e^{3x}v + 6e^{3x}v' + e^{3x}v'' これから方程式は

(x+1)(9e3xv+6e3xv′+e3xv′′)−(3x+4)(3e3xv+e3xv′)+3e3xv=(3x+2)e3x(x+1)(9e^{3x}v + 6e^{3x}v' + e^{3x}v'') - (3x+4)(3e^{3x}v + e^{3x}v') + 3e^{3x}v = (3x+2)e^{3x}

(x+1)v′′+(3x+2)v′=3x+2(x+1)v'' + (3x+2)v' = 3x+2 v′=Pv' = P とおくと

(x+1)P′+(3x+2)P=3x+2(x+1)P' + (3x+2)P = 3x+2 P′+3x+2x+1P=3x+2x+1P' + \frac{3x+2}{x+1}P = \frac{3x+2}{x+1}

α=3x+2x+1\alpha = \frac{3x+2}{x+1}、β=3x+2x+1\beta = \frac{3x+2}{x+1} とおくと ∫αdx=∫(3−1x+1)dx=3x−log⁡(x+1)\int \alpha dx = \int (3 - \frac{1}{x+1})dx = 3x - \log(x+1) これから

P=(x+1)e3x{∫e3xx+1⋅3x+2x+1dx+C1}=(x+1)e3x{∫(3x+1−1(x+1)2)e3xdx+C1}P = (x+1)e^{3x}\{\int \frac{e^{3x}}{x+1} \cdot \frac{3x+2}{x+1}dx + C_1\} = (x+1)e^{3x}\{\int (\frac{3}{x+1} - \frac{1}{(x+1)^2})e^{3x}dx + C_1\}

=(x+1)e3x{3⋅13⋅e3xx+1−∫−1(x+1)2e3xdx−∫1(x+1)2e3xdx+C1}= (x+1)e^{3x}\{3 \cdot \frac{1}{3} \cdot \frac{e^{3x}}{x+1} - \int \frac{-1}{(x+1)^2}e^{3x}dx - \int \frac{1}{(x+1)^2}e^{3x}dx + C_1\}

=(x+1)e3x{1x+1e3x+C1}=1+C1(x+1)e−3x= (x+1)e^{3x}\{\frac{1}{x+1}e^{3x} + C_1\} = 1 + C_1(x+1)e^{-3x} 元に戻して

v′=1+C1(x+1)e−3xv' = 1 + C_1(x+1)e^{-3x}

v=x+C1{−13(x+1)e−3x+13∫e−3xdx+C2}=x−C19{3(x+1)+1}e−3x+C1C2v = x + C_1\{-\frac{1}{3}(x+1)e^{-3x} + \frac{1}{3}\int e^{-3x}dx + C_2\} = x - \frac{C_1}{9}\{3(x+1) + 1\}e^{-3x} + C_1C_2

=x−C19(3x+4)e−3x+C1C2= x - \frac{C_1}{9}(3x+4)e^{-3x} + C_1C_2

y=e3xy = e^{3x}から

y=xe3x−C19(3x+4)+C1C2e3xy = xe^{3x} - \frac{C_1}{9}(3x+4) + C_1C_2e^{3x} −C19=A-\frac{C_1}{9} = A、C1C2=BC_1C_2 = Bと置くと

y=A(3x+4)+Be3x+xe3xy = A(3x+4) + Be^{3x} + xe^{3x} (A,Bは任意定数)(A,B \text{は任意定数})

解説: r31bn1z