高専の数学解説

第9章 9.12

(1)

(x+y)2dydx=4(x+y)^2 \frac{dy}{dx} = 4

x+y=ux+y=u とおくと dydx=dudx−1\dfrac{dy}{dx} = \dfrac{du}{dx} - 1 これから

u2(dudx−1)=4dudx=4+u2u2u^2\left(\frac{du}{dx} - 1\right) = 4 \qquad \frac{du}{dx} = \frac{4+u^2}{u^2}

∫u24+u2 du=∫dx∫(1−44+u2)du=∫dxu−2tan⁡−1u2=x+C元に戻して\int \frac{u^2}{4+u^2}\,du = \int dx \qquad \int\left(1 - \frac{4}{4+u^2}\right)du = \int dx \qquad u - 2\tan^{-1}\frac{u}{2} = x + C \qquad \text{元に戻して}

x+y−2tan⁡−1x+y2=x+Cx + y - 2\tan^{-1}\frac{x+y}{2} = x + C

y=2tan⁡−1x+y2+C(C は任意定数)y = 2\tan^{-1}\frac{x+y}{2} + C \qquad (\text{C \text{は任意定数}})

(2)

dydx=2x−4y+3x−2y−1x−2y−1=u とおくとdydx=12(1−dudx)これから\frac{dy}{dx} = \frac{2x-4y+3}{x-2y-1} \qquad x-2y-1=u \text{ とおくと} \qquad \frac{dy}{dx} = \frac{1}{2}\left(1-\frac{du}{dx}\right) \qquad \text{これから}

12(1−dudx)=2u+5u1−dudx=4u+10ududx=−3u−10u\frac{1}{2}\left(1-\frac{du}{dx}\right) = \frac{2u+5}{u} \qquad 1 - \frac{du}{dx} = \frac{4u+10}{u} \qquad \frac{du}{dx} = \frac{-3u-10}{u}

−∫u3u+10 du=∫dx−∫(13−103⋅110+3u)du=∫dx-\int \frac{u}{3u+10}\,du = \int dx \qquad -\int\left(\frac{1}{3} - \frac{10}{3}\cdot\frac{1}{10+3u}\right)du = \int dx

−13u+109log⁡(10+3u)=x+C19C1=C2 とおくと-\frac{1}{3}u + \frac{10}{9}\log(10+3u) = x + C_1 \qquad 9C_1 = C_2 \text{ とおくと}

−3u+10log⁡(3u+10)=9x+C2元に戻して-3u + 10\log(3u+10) = 9x + C_2 \qquad \text{元に戻して}

−3x+6y+3+10log⁡(3x−6y−3+10)=9x+C2−12(C2−3)=C とおくと-3x+6y+3+10\log(3x-6y-3+10) = 9x + C_2 \qquad -\frac{1}{2}(C_2-3) = C \text{ とおくと}

6x−3y−5log⁡(3x−6y+7)=C(C は任意定数)6x - 3y - 5\log(3x-6y+7) = C \qquad (\text{C \text{は任意定数}})

(3)

dydx+yx=(3x+2)xyxy=u とおくとy+xdydx=dudx\frac{dy}{dx} + \frac{y}{x} = (3x+2)xy \qquad xy = u \text{ とおくと} \qquad y + x\frac{dy}{dx} = \frac{du}{dx}

与式を整理して xdydx+y=(3x+2)x2yx\dfrac{dy}{dx} + y = (3x+2)x^2 y これから

dudx=ux(3x+2)\frac{du}{dx} = ux(3x+2)

∫1udu=∫x(3x+2)dxlog⁡u=x3+x2+C元に戻して\int \frac{1}{u} du = \int x(3x+2) dx \qquad \log u = x^3 + x^2 + C \qquad \text{元に戻して}

log⁡xy=x3+x2+C(C は任意定数)\log xy = x^3 + x^2 + C \qquad (\text{C \text{は任意定数}})

(4)

xdydx−y=y2−x2x2−1y=xu とおくとdydx=u+xdudxこれからx\frac{dy}{dx} - y = \frac{y^2 - x^2}{x^2 - 1} \qquad y = xu \text{ とおくと} \qquad \frac{dy}{dx} = u + x\frac{du}{dx} \qquad \text{これから}

x(u+xdudx)−xu=x2u2−x2x2−1dudx=u2−1x2−1x\left(u + x\frac{du}{dx}\right) - xu = \frac{x^2 u^2 - x^2}{x^2 - 1} \qquad \frac{du}{dx} = \frac{u^2 - 1}{x^2 - 1}

∫1u2−1du=∫1x2−1dx12log⁡u−1u+1=12(log⁡x−1x+1+C1)eC1=C とおくと\int \frac{1}{u^2 - 1} du = \int \frac{1}{x^2 - 1} dx \qquad \frac{1}{2}\log\frac{u-1}{u+1} = \frac{1}{2}\left(\log\frac{x-1}{x+1} + C_1\right) \qquad e^{C_1} = C \text{ とおくと}

(u−1)(x+1)(u+1)(x−1)=C(u−1)(x+1)=C(u+1)(x−1)元に戻して\frac{(u-1)(x+1)}{(u+1)(x-1)} = C \qquad (u-1)(x+1) = C(u+1)(x-1) \qquad \text{元に戻して}

(yx−1)(x+1)=C(yx+1)(x−1)(y−x)(x+1)=C(y+x)(x−1)(C は任意定数)\left(\frac{y}{x} - 1\right)(x+1) = C\left(\frac{y}{x} + 1\right)(x-1) \qquad (y-x)(x+1) = C(y+x)(x-1) \qquad (\text{C \text{は任意定数}})

解説: r31bn1z