(t+ty^2)dt+e^{t^2}ydy=0

Lösung

(t + t y^{2}) d t + e^{t^{2}} y d y = 0

y = e^{e^{- t^{2}} + c_{1}} - 1, y = - e^{e^{- t^{2}} + c_{1}} - 1

Schritte zur Lösung

(t + t y^{2}) d t + e^{t^{2}} y d y = 0

Solve separable ODE:

y = e^{e^{- t^{2}} + c_{1}} - 1, y = - e^{e^{- t^{2}} + c_{1}} - 1

y = e^{e^{- t^{2}} + c_{1}} - 1, y = - e^{e^{- t^{2}} + c_{1}} - 1

y^{^{'}} sec (t) = \frac{- y ^{11}}{10}

y^{^{'}} + y cos (x) = cos (x)

\frac{d y}{d x} = e^{x - 2 y}

(16 - x^{2}) \frac{d y}{d x} = 8 y

y^{^{'}} - \frac{1}{2 x} y = 0