y^'=ty

Lösung

y^{^{'}} = t y

y = e^{\frac{t ^{2}}{2} + c_{1}}

Schritte zur Lösung

y^{'} (t) = t y

Solve separable ODE:

y = e^{\frac{t ^{2}}{2} + c_{1}}

y = e^{\frac{t ^{2}}{2} + c_{1}}

\frac{\partial}{\partial x} ((ln (1 + e^{(x)}))^{2})

t \to 0 lim (\frac{2 e ^{t} - 2}{t})

\int 2 + 2 cos (x) d x

\int \frac{1}{3 - 4 x - x ^{2}} d x

d y = (x - 2 y cot (2 x)) d x