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Beliebt Rechnen >

(dy)/(dx)=3y+y^2-2

Lösung

\frac{d y}{d x} = 3 y + y^{2} - 2

Lösung

y = \frac{3 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} - \frac{3 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )}

Schritte zur Lösung

\frac{d y}{d x} = 3 y + y^{2} - 2

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

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

Löse nach

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

(\frac{1}{x ^{^{^{'}}}}) = 3 y + y^{2} - 2

Isoliere

x^{^{'}} : x^{^{'}} = \frac{1}{3 y + y ^{2} - 2}

Wenn

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

dann

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

x = \int \frac{1}{3 y + y ^{2} - 2} d y

\int \frac{1}{3 y + y ^{2} - 2} d y = - \frac{1}{17} (ln (\frac{2 y + 3}{17} + 1) - ln (\frac{2 y + 3}{17} - 1)) + c_{1}

x = - \frac{1}{17} (ln (\frac{2 y + 3}{17} + 1) - ln (\frac{2 y + 3}{17} - 1)) + c_{1}

Isoliere

y : y = \frac{3 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} - \frac{3 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )}

y = \frac{3 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 c_{1}}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} - \frac{3 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )} + \frac{17 e ^{17 x}}{2 ( e ^{17 x} - e ^{17 c_{1}} )}

Graph

Beliebte Beispiele

y^{^{'}} + (x + 8) y = 0

sec^2(x)tan(y)dx+sec^2(y)tan(x)dy=0

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

\frac{y}{x} y^{^{'}} + e^{y} = 0

(dy)/(dx)=2e^{x-y},y(1)=ln(2e+1)

\frac{d y}{d x} = 2 e^{x - y}, y (1) = ln (2 e + 1)

e^{3x}y-e^{5x}y^3y^'=0

e^{3 x} y - e^{5 x} y^{3} y^{^{'}} = 0