y^'x+y=0

Lösung

y^{^{'}} x + y = 0

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

Schritte zur Lösung

y^{'} (x) x + y = 0

Solve separable ODE:

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

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

y^{^{'}} - 5 y = 20, y (0) = 0

tan (x) d y - csc (y) d x = 0

6 y^{^{'}} + 3 t^{2} y = 12 t^{2}

\frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} (25)

y y^{^{'}} - 2 e^{x} = 0, y (0) = 8