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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

\frac{d}{d x} (3 lo g_{10} (x))

\frac{\partial}{\partial x} (\frac{1}{2 x + 4 y ^{2}})

\int_{0}^{3} x^{2} 9 - x^{2} d x

n = 0 \sum \infty (\frac{n x}{6})^{n}

x \to + 2 lim (2 - x)