x^2y^'+xy=y

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

u^{^{'}} = e^{9 x - H (x)}

y^{^{'}} = y (5 x^{2} + 3)

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

(t^{2} - 16 t + 48) (\frac{d y}{d t}) = y

\frac{d z}{d t} = 2 t e^{6 z}