(dy)/(dt)=((2y^4+t^4))/(ty^3)

Lösung

\frac{d y}{d t} = \frac{( 2 y ^{4} + t ^{4} )}{t y ^{3}}

y = t 4 - 1 + c_{1} t^{4}, y = - t 4 - 1 + c_{1} t^{4}

Schritte zur Lösung

\frac{d y}{d t} = \frac{( 2 y ^{4} + t ^{4} )}{t y ^{3}}

Solve Bernoulli ODE:

y = t 4 - 1 + c_{1} t^{4}, y = - t 4 - 1 + c_{1} t^{4}

y = t 4 - 1 + c_{1} t^{4}, y = - t 4 - 1 + c_{1} t^{4}

\frac{d y}{d t} + 20 \frac{6}{5} - \frac{6}{5} e^{- 20 t} = 24

\int \frac{tan ^{2} ( x )}{sec ( x )} d x

x \to 0 lim (\frac{e ^{9 x} - 1}{x})

\frac{d y}{d x} + \frac{3}{x} y = 2 x + 5

in v erse l a pl a ce \frac{x}{( x + 2 ) ( x ^{2} + 4 )}