第10章 10.18

(1)

$y = x^n$ として $y' = nx^{n-1}$ $y'' = n(n-1)x^{n-2}$

$x^2(x+1)n(n-1)x^{n-2} - 2x^2(nx^{n-1}) + 2(x-1)x^n = 0$

$(x^3 + x^2)(n^2 - n)x^{n-2} - 2nx^{n+1} + 2(x-1)x^n = 0$

$(n^2 - n)(x^{n+1} + x^n) - 2nx^{n+1} + 2(x^{n+1} - x^n) = 0$

$x^{n+1}(n^2 - 3n + 2) - 2x^n + (n^2 - n)x^n = 0$

これから $n \leq 2$ と考えて $n = 2$ なら $y' = 2x$ $y'' = 2$

$(x^3 + x^2)2 - 4x^3 + 2(x-1)x^2 = 0$

よって $n = 2$

(2)

$y = x^2u$ とおくと $y' = 2xu + x^2u'$ $y'' = 2u + 4xu' + x^2u''$

$x^2(x+1)(2u + 4xu' + x^2u'') - 2x^2(2xu + x^2u') + 2(x-1)x^2u = 0$

$x^4(x+1)u'' + x^3(4x + 4 - 2x)u' + x^2(2x + 2 - 4x + 2x - 2)u = 0$

$x(x+1)u'' + (2x + 4)u' = 0$

(3)

$u' = P$ とおくと $x(x+1)P' + (2x + 4)P = 0$

$\int \frac{1}{P}dP = -\int \frac{2x+4}{x(x+1)}dx = -\int \left(\frac{4}{x} - \frac{2}{x+1}\right)dx$ $\log P = -\log \frac{x^4}{(x+1)^2} = C_2$

元に戻して

$u' = C_2 \frac{(x+1)^2}{x^4}$

$\int du = C_2 \int \frac{(x+1)^2}{x^4}dx = C_2 \int \left(\frac{1}{x^2} + \frac{2}{x^3} + \frac{1}{x^4}\right)dx$ $u = C_2\left[-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{3x^3}\right] + C_3$ 元に戻して

$y = x^2\left\{-C_2\left(\frac{1}{x} - \frac{1}{x^2} - \frac{1}{3x^2}\right) + C_3\right\}$ $-C_2 = A$，$C_3 = B$ として

$y = A\left(1 + x + \frac{1}{3x}\right) + Bx^2$ （$A$，$B$ は任意定数）