y^'+ty=y

Lösung

y^{^{'}} + t y = y

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

Schritte zur Lösung

y^{'} (t) + t y = y

Solve separable ODE:

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

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

x^{2} \frac{d y}{d x} + x y - y = 0

u^{^{'}} = csc (H (x)) cos (\frac{x}{2})

\frac{d y}{d x} = - \frac{3}{2} y + \frac{1}{2} y^{2}

a d a + b d b = 0

y^{^{'}} = 5 y + 2