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

Lösung

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

y = 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4, y = - 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4

Schritte zur Lösung

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

Solve separable ODE:

y = 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4, y = - 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4

y = 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4, y = - 4 e^{c_{1} + \frac{4 x ^{5}}{5}} - 4

t an g e n t x^{3} + 2 x^{2} + 4

\int \frac{4 ^{x}}{x} d x

y^{^{''}} + 9 y = 4 sin (x)

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

d er i v a t i v e f (x) = \frac{cos ( x )}{1 + a sin ( x )}