y^'+xy=xy^{-1/2}

Lösung

y^{^{'}} + x y = x y^{- \frac{1}{2}}

y = 3 e^{\frac{- 3 x ^{2} - 6 c _{1}}{2}} - 2 e^{- \frac{3 x ^{2}}{4} - \frac{3 c _{1}}{2}} + 1

Schritte zur Lösung

y^{'} (x) + x y = x y^{- \frac{1}{2}}

Solve separable ODE:

y = 3 e^{\frac{- 3 x ^{2} - 6 c _{1}}{2}} - 2 e^{- \frac{3 x ^{2}}{4} - \frac{3 c _{1}}{2}} + 1

y = 3 e^{\frac{- 3 x ^{2} - 6 c _{1}}{2}} - 2 e^{- \frac{3 x ^{2}}{4} - \frac{3 c _{1}}{2}} + 1

\frac{d z}{d t} + 2 e^{t + z} = 0

\frac{d y}{d x} = 3 x^{2} y, y (1) = 1

2 x y d x + y^{2} d y = 0

x 1 - y^{2} d x + y 1 - x^{2} d y = 0

\frac{d v}{d t} = g - k v (t)