(dy)/(dx)=sin(x)(1-cos(x))(y^2-4)

Lösung

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

y = \frac{2 e ^{4 c_{1}}}{e ^{2 (c o s^{2} (x) - 2 c o s (x) + 1)} - e ^{4 c_{1}}} + \frac{2 e ^{2 c o s^{2} (x) - 4 c o s (x) + 2}}{e ^{2 (c o s^{2} (x) - 2 c o s (x) + 1)} - e ^{4 c_{1}}}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{2 e ^{4 c_{1}}}{e ^{2 (c o s^{2} (x) - 2 c o s (x) + 1)} - e ^{4 c_{1}}} + \frac{2 e ^{2 c o s^{2} (x) - 4 c o s (x) + 2}}{e ^{2 (c o s^{2} (x) - 2 c o s (x) + 1)} - e ^{4 c_{1}}}

y = \frac{2 e ^{4 c_{1}}}{e ^{2 (c o s^{2} (x) - 2 c o s (x) + 1)} - e ^{4 c_{1}}} + \frac{2 e ^{2 c o s^{2} (x) - 4 c o s (x) + 2}}{e ^{2 (c o s^{2} (x) - 2 c o s (x) + 1)} - e ^{4 c_{1}}}

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

\frac{d P}{d t} = - 3 P

y^{^{'}} = x y + 8 x + 5 y + 40

y^{^{'}} = 2 x^{3} \cdot y^{2}

- y^{^{'}} - 9 y = 0