y^'=10te^{-y}

Lösung

y^{^{'}} = 10 t e^{- y}

y = ln (5 t^{2} + c_{1})

Schritte zur Lösung

y^{'} (t) = 10 t e^{- y}

Solve separable ODE:

y = ln (5 t^{2} + c_{1})

y = ln (5 t^{2} + c_{1})

x^{5} d y = - y^{5} d x

\frac{d x}{d t} = \frac{4}{x e ^{t + 6 x}}

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

\frac{d}{d x} (4 x^{3} + 2 y^{5}) = 6

t^{2} y + 11 t y^{^{'}} - 11 y = 0, y^{1} (0) = t