y^{''}+4y=cos(x),y(0)=6,y^'(0)=-6

Lösung

y^{^{''}} + 4 y = cos (x), y (0) = 6, y^{^{'}} (0) = - 6

y = \frac{17}{3} cos (2 x) - 3 sin (2 x) + \frac{1}{3} cos (x)

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{17}{3} cos (2 x) - 3 sin (2 x) + \frac{1}{3} cos (x)

y = \frac{17}{3} cos (2 x) - 3 sin (2 x) + \frac{1}{3} cos (x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{2 x}

y^{^{'''}} - y^{^{'}} = 3 t

y^{^{'''}} - y^{^{'}} = 5 t

4 y^{^{''}} - y = e^{\frac{x}{2}} + 6

y^{^{''}} + y^{^{'}} = x^{2}