y^'+tan(3x)y=0,y(0)=5

Lösung

y^{^{'}} + tan (3 x) y = 0, y (0) = 5

y = cos^{\frac{1}{3}} (3 x) \cdot 5

Schritte zur Lösung

y^{'} (x) + tan (3 x) y = 0

Solve separable ODE:

y = cos^{\frac{1}{3}} (3 x) \cdot 5

y = cos^{\frac{1}{3}} (3 x) \cdot 5

2 x y y^{^{'}} - 2 x^{2} + 1 = 0!

\frac{d y}{d x} = x y, y \neq = 0

y^{^{'}} = \frac{20 - 0.06 y}{0.02}

y^{^{'}} = (x^{4} + 7) (cos (y))^{2}

y^{^{'}} = r \cdot y \cdot (1 - \frac{y}{k})