y^'=2tycos(t^2)

Lösung

y^{^{'}} = 2 t y cos (t^{2})

y = e^{s i n (t^{2}) + c_{1}}

Schritte zur Lösung

y^{'} (t) = 2 t y cos (t^{2})

Solve separable ODE:

y = e^{s i n (t^{2}) + c_{1}}

y = e^{s i n (t^{2}) + c_{1}}

y^{^{'}} sec (x) = 4 y^{2}, y (\frac{5 π}{2}) = \frac{1}{5}

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

y^{^{'}} = \frac{- ( 3 y ^{2} + 1 ) x}{x ^{2} + 1}, y (0) = 1

y \cdot y^{^{'}} = x

\frac{d y}{d t} = y - y t^{2}