y^{''}+2y^'+5y=4e^{-t}cos(2t)

Lösung

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{- t} cos (2 t)

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{- t} t sin (2 t)

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 5 y = 4 e^{- t} cos (2 t)

Solve linear ODE:

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{- t} t sin (2 t)

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{- t} t sin (2 t)

x \cdot d x - y^{2} \cdot d y = 0

y^{^{'}} = x^{2} + 1

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{15}{2}}} d t

\int (x^{6} - 2 x^{3})^{5} (x^{5} - x^{2}) d x

d er i v a t i v e \frac{1}{3 ( 8 - x ^{3} ) ^{8}}