yy^'=[sin(x)-cos(x)]sqrt(1-y^2)

Lösung

y y^{^{'}} = [sin (x) - cos (x)] 1 - y^{2}

y = 2 c_{1} cos (x) - 2 sin (x) cos (x) - cos^{2} (x) - c_{1}^{2} - sin^{2} (x) + 2 c_{1} sin (x) + 1, y = - 2 c_{1} cos (x) - 2 sin (x) cos (x) - cos^{2} (x) - c_{1}^{2} - sin^{2} (x) + 2 c_{1} sin (x) + 1

Schritte zur Lösung

y y^{'} (x) = [sin (x) - cos (x)] 1 - y^{2}

Solve separable ODE:

y = 2 c_{1} cos (x) - 2 sin (x) cos (x) - cos^{2} (x) - c_{1}^{2} - sin^{2} (x) + 2 c_{1} sin (x) + 1, y = - 2 c_{1} cos (x) - 2 sin (x) cos (x) - cos^{2} (x) - c_{1}^{2} - sin^{2} (x) + 2 c_{1} sin (x) + 1

y = 2 c_{1} cos (x) - 2 sin (x) cos (x) - cos^{2} (x) - c_{1}^{2} - sin^{2} (x) + 2 c_{1} sin (x) + 1, y = - 2 c_{1} cos (x) - 2 sin (x) cos (x) - cos^{2} (x) - c_{1}^{2} - sin^{2} (x) + 2 c_{1} sin (x) + 1

\frac{d y}{d x} = 13 x sec (y)

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

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

\frac{d y}{d x} = \frac{x ^{9}}{y ^{6}}

\frac{d y}{d x} = - \frac{x}{y}, y (0) = 4