高専の数学解説

第9章 9.15

(1)

xy′+y=1−x2xy' + y = 1 - x^2

y′+1xy=1x−xy' + \frac{1}{x}y = \frac{1}{x} - x

P=1xP = \frac{1}{x}、Q=1x−xQ = \frac{1}{x} - x とおくと ∫Pdx=log⁡x\int Pdx = \log x これから

y=e−log⁡x[∫elog⁡x(1x−x)dx+C]=1x∫(1−x2)dx+C]=1x[x−13x3+C]y = e^{-\log x}\left[\int e^{\log x}\left(\frac{1}{x} - x\right)dx + C\right] = \frac{1}{x}\int (1 - x^2)dx + C] = \frac{1}{x}\left[x - \frac{1}{3}x^3 + C\right]

=1−13x2+Cx= 1 - \frac{1}{3}x^2 + \frac{C}{x}（CC は任意定数）

(2)

(1+x2)y′=xy+1(1 + x^2)y' = xy + 1 y′=x1+x2y=11+x2y' = \frac{x}{1 + x^2}y = \frac{1}{1 + x^2}

P=−xx2+1P = -\frac{x}{x^2 + 1}、Q=11+x2Q = \frac{1}{1 + x^2} とおくと ∫Pdx=−12log⁡(x2+1)=log⁡(x2+1)−12\int Pdx = -\frac{1}{2}\log(x^2 + 1) = \log(x^2 + 1)^{-\frac{1}{2}} これから

y=x2+1[∫(1x2+1−1x2+1)dx+C]y = \sqrt{x^2 + 1}\left[\int \left(\frac{1}{\sqrt{x^2 + 1}} - \frac{1}{x^2 + 1}\right)dx + C\right]

∫1x2+1−1x2+1dxx=tan⁡θ とおくとdx=1cos⁡2θdθ これから\int \frac{1}{\sqrt{x^2 + 1}} - \frac{1}{x^2 + 1}dx \quad x = \tan \theta \text{ とおくと} \quad dx = \frac{1}{\cos^2 \theta}d\theta \text{ これから}

=∫1sin⁡2θ+cos⁡2θcos⁡2θ⋅1sin⁡2θ+cos⁡2θcos⁡2θdθ=∫cos⁡θdθ=sin⁡θ元に戻して= \int \frac{1}{\sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta}}} \cdot \frac{1}{\frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta}}d\theta = \int \cos \theta d\theta = \sin \theta \quad \text{元に戻して}

=sin⁡(tan⁡−1x)→xx2+1 これから= \sin(\tan^{-1}x) \to \frac{x}{\sqrt{x^2 + 1}} \text{ これから}

y=x2+1[xx2+1+C]=x+Cx2+1y = \sqrt{x^2 + 1}\left[\frac{x}{\sqrt{x^2 + 1}} + C\right] = x + C\sqrt{x^2 + 1}（CC は任意定数）

(3)

xy′+y=log⁡xxy' + y = \log x y′+1xy=1xlog⁡xy' + \frac{1}{x}y = \frac{1}{x}\log x

P=1xP = \frac{1}{x}，Q=1xlog⁡xQ = \frac{1}{x}\log x とおくと ∫Pdx=log⁡x\int Pdx = \log x これから

y=1x{∫(x⋅1xdx+C)}=1x(xlog⁡x−x+C)=log⁡x−1+Cxy = \frac{1}{x}\left\{\int\left(x \cdot \frac{1}{x} dx + C\right)\right\} = \frac{1}{x}(x \log x - x + C) = \log x - 1 + \frac{C}{x} （CC は任意定数）

(4)

y′cos⁡x+ysin⁡x=sin⁡x⋅cos⁡2xy'\cos x + y \sin x = \sin x \cdot \cos^2 x y′+y⋅sin⁡xcos⁡x=sin⁡x⋅cos⁡xy' + y \cdot \frac{\sin x}{\cos x} = \sin x \cdot \cos x

P=sin⁡xcos⁡xP = \frac{\sin x}{\cos x}，Q=sin⁡x⋅cos⁡xQ = \sin x \cdot \cos x とおくと ∫Pdx=∫sin⁡xcos⁡xdx\int Pdx = \int \frac{\sin x}{\cos x} dx cos⁡x=t\cos x = t とおくと

∫Pdx=∫(−1t)dt=−log⁡t\int Pdx = \int \left(-\frac{1}{t}\right) dt = -\log t 元に戻して ∫Pdx=−log⁡(cos⁡x)\int Pdx = -\log(\cos x) これから

y=cos⁡x{∫sin⁡x⋅cos⁡xcos⁡xdx+C}=cos⁡x{∫sin⁡xdx+C}=cos⁡x(−cos⁡x+C)y = \cos x\left\{\int \frac{\sin x \cdot \cos x}{\cos x} dx + C\right\} = \cos x\{\int \sin x dx + C\} = \cos x(-\cos x + C)

=Ccos⁡x−cos⁡2x= C \cos x - \cos^2 x （CC は任意定数）

(5)

(2x+1)y′−x+y=0(2x + 1)y' - x + y = 0 y′+x2x+1y=x2x+1y' + \frac{x}{2x+1}y = \frac{x}{2x+1}

P=12x+1P = \frac{1}{2x+1}，Q=x2x+1Q = \frac{x}{2x+1} とおくと ∫Pdx=12log⁡(2x+1)\int Pdx = \frac{1}{2}\log(2x+1) これから

y=12x+1{∫2x+1⋅x2x+1dx+C}=12x+1⋅12[∫2x+1−12x+1dx+C]y = \frac{1}{\sqrt{2x+1}}\left\{\int\sqrt{2x+1} \cdot \frac{x}{2x+1} dx + C\right\} = \frac{1}{\sqrt{2x+1}} \cdot \frac{1}{2}\left[\int\sqrt{2x+1} - \frac{1}{2x+1} dx + C\right]

=12x+1⋅12[13(2x+1)3/2−2x+1]+C}=16(2x+1)−12+C2x+1= \frac{1}{\sqrt{2x+1}} \cdot \frac{1}{2}\left[\frac{1}{3}(2x+1)^{3/2} - \sqrt{2x+1}\right] + C\} = \frac{1}{6}(2x+1) - \frac{1}{2} + \frac{C}{\sqrt{2x+1}}

=x−13+C2x+1= \frac{x-1}{3} + \frac{C}{\sqrt{2x+1}} （CC は任意定数）

(6)

(1+x2)y′+xy=0(1+x^2)y'+xy = 0

y′+x1+x2y=11+x2y'+\frac{x}{1+x^2}y = \frac{1}{1+x^2}

P=x1+x2,Q=11+x2 とおくと ∫Pdx=12log⁡(1+x2)P=\frac{x}{1+x^2}, \quad Q=\frac{1}{1+x^2} \text{ とおくと } \int Pdx = \frac{1}{2}\log(1+x^2)

y=11+x2{∫1+x21+x2dx+C}=11+x2{∫11+x2dx+C}y = \frac{1}{\sqrt{1+x^2}}\left\{\int\frac{\sqrt{1+x^2}}{1+x^2}dx+C\right\} = \frac{1}{\sqrt{1+x^2}}\left\{\int\frac{1}{\sqrt{1+x^2}}dx+C\right\}

=11+x2(log⁡∣x+x2+1∣+C)（C は任意定数）= \frac{1}{\sqrt{1+x^2}}\left(\log|x+\sqrt{x^2+1}|+C\right) \quad \text{（C \text{は任意定数}）}

解説: r31bn1z