xy^'+y=1

Lösung

x y^{^{'}} + y = 1

y = - \frac{c _{1}}{x} + 1

Schritte zur Lösung

x y^{'} (x) + y = 1

Solve separable ODE:

y = - \frac{c _{1}}{x} + 1

y = - \frac{c _{1}}{x} + 1

3 t \frac{d y}{d t} = y cos (t)

y^{^{'}} = 4 x y^{2}

\frac{d y}{d x} - 3 x^{2} y = 0

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

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