y^{''}+2y^'+2y=sin(kx)

Lösung

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (k x)

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

Schritte zur Lösung

y^{''} (x) + 2 y^{'} (x) + 2 y = sin (k x)

Solve linear ODE:

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

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

\int sec (2 x) tan (2 x) - \frac{4}{x ^{2}} d x

\int (cos (y)) d y

t a y l or \frac{8}{1 - x}

\int \frac{x ^{5} - x ^{3} + 2 x}{x ^{4}} d x

x \to 5 lim (x - 4)