第10章 10.2

(1)

y′=ex−xy' = e^x - x

y′=Py' = P とおくと P′=ex−xP' = e^x - x dPdx=ex−x\frac{dP}{dx} = e^x - x

∫dP=∫(ex−x)dx\int dP = \int (e^x - x)dx P=ex−12x2+C1P = e^x - \frac{1}{2}x^2 + C_1 元に戻して

dydx=ex−12x2+C1\frac{dy}{dx} = e^x - \frac{1}{2}x^2 + C_1

∫dy=∫(ex−12x2+C1)dx\int dy = \int (e^x - \frac{1}{2}x^2 + C_1)dx y=ex−16x3+C1x+C2y = e^x - \frac{1}{6}x^3 + C_1 x + C_2 C1=A,C2=BC_1 = A, C_2 = B とおくと

y=ex−16x3+Ax+By = e^x - \frac{1}{6}x^3 + Ax + B (A,B は任意定数)

(2)

y′′=xcos⁡xy'' = x\cos x

y′=Py' = P とおくと P′=xcos⁡xP' = x\cos x dPdx=xcos⁡x\frac{dP}{dx} = x\cos x

∫dP=∫xcos⁡xdx\int dP = \int x\cos x dx P=xsin⁡x−∫sin⁡xdx=xsin⁡x+cos⁡x+C1P = x\sin x - \int \sin x dx = x\sin x + \cos x + C_1 元に戻して

dydx=xsin⁡x+cos⁡x+C1\frac{dy}{dx} = x\sin x + \cos x + C_1

∫dy=∫(xsin⁡x+cos⁡x+C1)dx\int dy = \int (x\sin x + \cos x + C_1)dx

y=−xcos⁡x+∫cos⁡xdx+sin⁡x+C1x+C2=−xcos⁡x+sin⁡x+sin⁡x+C1x+C2y = -x\cos x + \int \cos x dx + \sin x + C_1 x + C_2 = -x\cos x + \sin x + \sin x + C_1 x + C_2

=2sin⁡x−xcos⁡x+C1x+C2= 2\sin x - x\cos x + C_1 x + C_2 C1=A,C2=BC_1 = A, C_2 = B とおくと

y=2sin⁡x−xcos⁡x+Ax+By = 2\sin x - x\cos x + Ax + B (A,B は任意定数)

(3)

y′′=1−xlog⁡xy''=1-x\log x

y′=Py'=P とおくと xP′=1−xlog⁡xxP'=1-x\log x dPdx=1x−log⁡x\frac{dP}{dx}=\frac{1}{x}-\log x

∫dP=∫(1x−log⁡x)dxP=log⁡x−xlog⁡x+x+C1元に戻して\int dP = \int \left(\frac{1}{x}-\log x\right)dx P=\log x - x\log x + x + C_1 \text{元に戻して}

dPdx=log⁡x−xlog⁡x+x+C1\frac{dP}{dx}=\log x - x\log x + x + C_1

∫dy=∫(log⁡x−xlog⁡x+x+C1)dx\int dy = \int (\log x - x\log x + x + C_1)dx

y=xlog⁡x−x−12x2log⁡x+12∫xdx+C1x+C2(C1−1)=A,C2=B とおくとy = x\log x - x - \frac{1}{2}x^2\log x + \frac{1}{2}\int xdx + C_1 x + C_2 (C_1-1)=A, C_2=B \text{ とおくと}

y=xlog⁡x−x22log⁡x+34x2+Ax+B(A, B は任意定数)y = x\log x - \frac{x^2}{2}\log x + \frac{3}{4}x^2 + Ax + B \text{(A, B \text{は任意定数})}

(4)

y′′=x2e2xy''=x^2 e^{2x}

y′=Py'=P とおくと P′=x2e2xP'=x^2 e^{2x} dPdx=x2e2x\frac{dP}{dx}=x^2 e^{2x}

∫dP=∫x2e2xdx=12x2e2x−∫xe2xdx=12x2e2x−12xe2x+∫12e2xdx\int dP = \int x^2 e^{2x} dx = \frac{1}{2}x^2 e^{2x} - \int xe^{2x} dx = \frac{1}{2}x^2 e^{2x} - \frac{1}{2}xe^{2x} + \int \frac{1}{2}e^{2x}dx

=12x2e2x−12xe2x+14e2x+C1元に戻して= \frac{1}{2}x^2 e^{2x} - \frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} + C_1 \text{元に戻して}

dydx=12x2e2x−12xe2x+14e2x+C1\frac{dy}{dx}=\frac{1}{2}x^2 e^{2x} - \frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} + C_1

∫dy=∫(12x2e2x−12xe2x+14e2x+C1)dx\int dy = \int \left(\frac{1}{2}x^2 e^{2x} - \frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} + C_1\right)dx

y=12[12x2e2x−12xe2x+14e2x]−12[12xe2x−14e2x]+18e2x+C1x+C2y = \frac{1}{2}\left[\frac{1}{2}x^2 e^{2x} - \frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x}\right] - \frac{1}{2}\left[\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}\right] + \frac{1}{8}e^{2x} + C_1 x + C_2

C1=A,C2=BC_1 = A, C_2 = B とおくと

=18[2x2−4x+3]e2x+Ax+B(A, B は任意定数)= \frac{1}{8}[2x^2 - 4x + 3]e^{2x} + Ax + B \text{(A, B \text{は任意定数})}

解説: r31bn1z