3y^4-1+12xy^3y^'=0

Lösung

3 y^{4} - 1 + 12 x y^{3} y^{^{'}} = 0

y = 4 \frac{c _{1} + x}{3 x}, y = - 4 \frac{c _{1} + x}{3 x}

Schritte zur Lösung

3 y^{4} - 1 + 12 x y^{3} y^{'} (x) = 0

Solve separable ODE:

y = 4 \frac{c _{1} + x}{3 x}, y = - 4 \frac{c _{1} + x}{3 x}

y = 4 \frac{c _{1} + x}{3 x}, y = - 4 \frac{c _{1} + x}{3 x}

\frac{d y}{d x} = \frac{2 x ^{2}}{y + 1}

y^{^{'}} - y = - 1

y^{^{'}} = \frac{x + 4}{y ^{2}}

y^{^{'}} = \frac{4 x y}{x ^{2} + 1}

x y^{2} y^{^{'}} = x + x^{3}, y (1) = 2