y-4y^'+5y=0

Lösung

y - 4 y^{^{'}} + 5 y = 0

y = e^{\frac{3 t + c _{1}}{2}}

Schritte zur Lösung

y - 4 y^{'} (t) + 5 y = 0

Solve separable ODE:

y = e^{\frac{3 t + c _{1}}{2}}

y = e^{\frac{3 t + c _{1}}{2}}

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

y x + 12 x + (ln (x) - 3) y^{^{'}} = 0

y^{^{'}} = \frac{( y + 4 )}{x}

f (x) = \frac{x}{sin ( x )} f^{^{'}} (x)

\frac{d}{d θ} (ln (y (θ))) = \frac{θ - 8}{θ cos ( θ )}