x+x(y^')2-1=0

Lösung

x + x (y^{^{'}}) 2 - 1 = 0

y = \frac{1}{2} (- x + ln (x)) + c_{1}

Schritte zur Lösung

x + x (y^{'} (x)) 2 - 1 = 0

Isoliere

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

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

y = \int \frac{- x + 1}{2 x} d x

\int \frac{- x + 1}{2 x} d x = \frac{1}{2} (- x + ln (x)) + c_{1}

y = \frac{1}{2} (- x + ln (x)) + c_{1}

y^{^{''}} + 2 y^{^{'}} = 3

\frac{d}{d x} (y - 3 x)

\frac{d}{d x} (2 x^{3} + \frac{1}{x ^{2}} + 16 x^{\frac{7}{2}} + 5)

2 x y^{^{'}} + y = 6 x, y (4) = 20

x^{^{''''}} = 0