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

Lösung

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

y = - \frac{2}{3 x ^{2} + 2 c _{1}} - \frac{12 x ^{2}}{3 x ^{2} + 2 c _{1}} - \frac{8 c _{1}}{3 x ^{2} + 2 c _{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{2}{3 x ^{2} + 2 c _{1}} - \frac{12 x ^{2}}{3 x ^{2} + 2 c _{1}} - \frac{8 c _{1}}{3 x ^{2} + 2 c _{1}}

y = - \frac{2}{3 x ^{2} + 2 c _{1}} - \frac{12 x ^{2}}{3 x ^{2} + 2 c _{1}} - \frac{8 c _{1}}{3 x ^{2} + 2 c _{1}}

f (t) = t - 4

\int \frac{6 x ^{2} - 5 x + 8}{x ^{3} + 4 x} d x

\int_{0}^{4} 16 x^{2} - x^{4} d x

\frac{\partial}{\partial x} (x^{2} sin (6 t))

\frac{d}{d x} (- \frac{e ^{- x} ( x ^{3} + 3 ( x ^{2} + 2 x + 2 ) )}{x ^{4}})