y^'=((x^3+x))/(4y^3)

Lösung

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

y = \frac{4 c _{1} + x ^{4} + 2 x ^{2}}{2}, y = - \frac{4 c _{1} + x ^{4} + 2 x ^{2}}{2}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{4 c _{1} + x ^{4} + 2 x ^{2}}{2}, y = - \frac{4 c _{1} + x ^{4} + 2 x ^{2}}{2}

y = \frac{4 c _{1} + x ^{4} + 2 x ^{2}}{2}, y = - \frac{4 c _{1} + x ^{4} + 2 x ^{2}}{2}

y^{^{'}} = 0.002 \cdot (1000 - y), y (0) = 1

v = 50 x \frac{d v}{d x}

\frac{d x}{d t} = 18 - 0.72 \cdot x

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

\frac{d y}{d t} y = (4 t - 1) (3 t - 6)^{- 1}