4y^{''}+9y=0,y(0)=2,y^'(0)=3

Lösung

4 y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y = 2 cos (\frac{3 t}{2}) + 2 sin (\frac{3 t}{2})

Schritte zur Lösung

4 y^{''} (t) + 9 y = 0

Solve linear ODE:

y = 2 cos (\frac{3 t}{2}) + 2 sin (\frac{3 t}{2})

y = 2 cos (\frac{3 t}{2}) + 2 sin (\frac{3 t}{2})

\frac{d}{d x} (a (2 e^{- 3 x} x - 3 e^{- 3 x} x^{2}))

\frac{\partial}{\partial x} (- sin (4 x))

d er i v a t i v e f (x) = x^{2} + 2 x + 3

\int 4 \frac{x}{( x + 2 ) ^{(\frac{1}{4})}} d x

x \to 1 - lim (3^{\frac{1}{x}})