yy^'+2sin(2x)=0

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

y^{^{'}} = 120 - 2 y

y^{^{'}} = 135 x^{2} y + 18 x y + 30 x^{2} + 4 x

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

(x^{2} y^{2}) d x - \frac{1}{x y} d y = 0

2 x - 6 y^{2} \frac{d y}{d x} + 4 = 0