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

Lösung

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

y = x 3 12 ln (x) + c_{1}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = x 3 12 ln (x) + c_{1}

y = x 3 12 ln (x) + c_{1}

n = 0 \sum \infty \frac{1}{2 n !}

y^{^{'}} = cos^{2} (y) ln (x)

\frac{\partial}{\partial x} (e^{5 x} sin (5 y))

l o n g d i v i s i o n \frac{x ^{2} - 1}{x ^{2} + x + 1}

d er i v a t i v e y = - \frac{2}{x ^{2}}