dy=(y^2-4)dx

Lösung

d y = (y^{2} - 4) d x

y = \frac{2 e ^{4 c_{1}}}{e ^{4 x} - e ^{4 c_{1}}} + \frac{2 e ^{4 x}}{e ^{4 x} - e ^{4 c_{1}}}

Schritte zur Lösung

d y = (y^{2} - 4) d x

Lass

x

die abhängige Variable sein. Teile durch

d y

1 = (y^{2} - 4) \frac{d x}{d y}

Ersetze

\frac{d x}{d y}

mit

x^{^{'}}

1 = (y^{2} - 4) x^{^{^{'}}}

Isoliere

x^{^{'}} : x^{^{'}} = \frac{1}{y ^{2} - 4}

Wenn

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

dann

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

x = \int \frac{1}{y ^{2} - 4} d y

\int \frac{1}{y ^{2} - 4} d y = - \frac{1}{4} (ln (\frac{y}{2} + 1) - ln (\frac{y}{2} - 1)) + c_{1}

x = - \frac{1}{4} (ln (\frac{y}{2} + 1) - ln (\frac{y}{2} - 1)) + c_{1}

Isoliere

y : y = \frac{2 e ^{4 c_{1}}}{e ^{4 x} - e ^{4 c_{1}}} + \frac{2 e ^{4 x}}{e ^{4 x} - e ^{4 c_{1}}}

y = \frac{2 e ^{4 c_{1}}}{e ^{4 x} - e ^{4 c_{1}}} + \frac{2 e ^{4 x}}{e ^{4 x} - e ^{4 c_{1}}}

h^{^{'}} = \frac{2 h}{x ln ( x )}

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

\frac{d y}{d x} + 12 x^{2} y = 5 x^{2}, y (0) = 4

x \frac{d y}{d x} = e^{4 y + x}

y^{^{'}} = y^{2} x cos (x^{2})