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

Lösung

y^{^{'}} = \frac{x ^{2} + 3}{4 y ^{3}}

y = 4 \frac{x ^{3}}{3} + 3 x + c_{1}, y = - 4 \frac{x ^{3}}{3} + 3 x + c_{1}

Schritte zur Lösung

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

Solve separable ODE:

y = 4 \frac{x ^{3}}{3} + 3 x + c_{1}, y = - 4 \frac{x ^{3}}{3} + 3 x + c_{1}

y = 4 \frac{x ^{3}}{3} + 3 x + c_{1}, y = - 4 \frac{x ^{3}}{3} + 3 x + c_{1}

x y^{^{'}} - y = x y

d y = (y^{2} - 4) d x

h^{^{'}} = \frac{2 h}{x ln ( x )}

(y^{2} - 1) \frac{d x}{d y} = x

\frac{d y}{d x} + 12 x^{2} y = 5 x^{2}, y (0) = 4