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

Lösung

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

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

Schritte zur Lösung

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

\int \frac{8}{4 - x ^{2}} d x

\int \frac{x ^{2} - 1}{x ^{3} - 3 x + 1} d x

d er i v a t i v e f (x) = e^{- 0.5 x} (4 - x) - 2

n = 1 \sum \infty \frac{5}{4 n ^{2} - 1}

\int_{\frac{1}{5}}^{7} 8 x ln (5 x) d x