y^{''}+y= 1/(2k^2)(1+cos(2t))

Lösung

y^{^{''}} + y = \frac{1}{2 k ^{2}} (1 + cos (2 t))

y = c_{1} cos (t) + c_{2} sin (t) + \frac{cos ^{4} ( t ) + sin ( t ) ( 3 sin ( t ) - sin ^{3} ( t ) )}{3 k ^{2}}

Schritte zur Lösung

y^{''} (t) + y = \frac{1}{2 k ^{2}} (1 + cos (2 t))

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + \frac{cos ^{4} ( t ) + sin ( t ) ( 3 sin ( t ) - sin ^{3} ( t ) )}{3 k ^{2}}

y = c_{1} cos (t) + c_{2} sin (t) + \frac{cos ^{4} ( t ) + sin ( t ) ( 3 sin ( t ) - sin ^{3} ( t ) )}{3 k ^{2}}

\frac{d}{d x} ((x^{5} + 1)^{4})

t an g e n t f (x) = \frac{1}{x}, at x = 81

t an g e n t arccos (x) + x, at x = 0

d er i v a t i v e y = 9 + 5 x^{2} - 2 x^{3}

2 y^{^{'}} + 4 y = 2 t e^{- 2 t}