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

解

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

y = 2 ln (x + 1) (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1) + c_{1} (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1)

解答ステップ

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

線形 ODE を解く:

y = 2 ln (x + 1) (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1) + c_{1} (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1)

y = 2 ln (x + 1) (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1) + c_{1} (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1)