(x-1)^3(dy)/(dx)+4(x-1)^2y=x+1

解答

(x - 1)^{3} \frac{d y}{d x} + 4 (x - 1)^{2} y = x + 1

y = \frac{x ^{3}}{3 ( x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1 )} - \frac{x}{x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1} + \frac{c _{1}}{x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1}

求解步骤

(x - 1)^{3} \frac{d y}{d x} + 4 (x - 1)^{2} y = x + 1

求解线性方程:

y = \frac{x ^{3}}{3 ( x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1 )} - \frac{x}{x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1} + \frac{c _{1}}{x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1}

y = \frac{x ^{3}}{3 ( x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1 )} - \frac{x}{x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1} + \frac{c _{1}}{x ^{4} - 4 x ^{3} + 6 x ^{2} - 4 x + 1}

y^{^{'}} = - \frac{1 + x + y}{1 + x}

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