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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

\frac{d}{d θ} (\frac{1}{4} sin^{2} (2 θ))

n = 0 \sum \infty e^{- \frac{3 n}{2}}

\int 4 x^{2} + 2 + \frac{5}{x ^{2} + 1} d x

\frac{\partial}{\partial t} (3 arcsin (t))

t an g e n t (x^{2} + y^{2})^{2} = 4 x^{2} y, (- 1, 1)