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

Lösung

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

y = - 2 x + c_{1} x^{3}, y = - - 2 x + c_{1} x^{3}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = - 2 x + c_{1} x^{3}, y = - - 2 x + c_{1} x^{3}

y = - 2 x + c_{1} x^{3}, y = - - 2 x + c_{1} x^{3}

\int \frac{x ^{2}}{- 4 x ^{2} - 12 x - 5} d x

x \to 8 lim (\frac{( 3 x - 2 )}{( x - 8 )})

f^{^{'}} (x) = \frac{( 5 x ^{2} - 9 x + 8 )}{7 x + 6}

\int x^{5} cos (2 x) d x

t an g e n t y = (1 + 4 x)^{\frac{3}{2}}, at x = 2