y^{''}+9y=2cos(3x),y(0)=1,y^'(0)=0

Lösung

y^{^{''}} + 9 y = 2 cos (3 x), y (0) = 1, y^{^{'}} (0) = 0

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

Schritte zur Lösung

y^{''} (x) + 9 y = 2 cos (3 x)

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (x^{2} \cdot e^{- 2 y})

\frac{\partial}{\partial x} (3 y^{2} x)

\frac{\partial}{\partial x} (6 x^{7 y})

\frac{d}{d x} (sin (x + sin (7 x)))

d er i v a t i v e y = (x - 1) e^{x}