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

解

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

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

解答ステップ

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

ベルヌーイ ODE を解く:

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

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