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

Lösung

2 y^{^{'}} + y = 0, y (0) = e^{- \frac{x}{2}}

y = e^{\frac{- x + c _{1}}{2}}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{\frac{- x + c _{1}}{2}}

y = e^{\frac{- x + c _{1}}{2}}

(x^{3} + x y^{2}) d x + (x^{2} y + y^{3}) d y = 0

\frac{d x}{d t} = 50 - \frac{x}{100}

y^{^{'}} = y y^{2} - 1

(x - 3) y^{^{'}} + 2 y = 0

\frac{d y}{d x} - \frac{y}{3 x - 1} = 0