y^{''}+y=3e^{-4x}

Lösung

y^{^{''}} + y = 3 e^{- 4 x}

y = c_{1} cos (x) + c_{2} sin (x) + \frac{3}{17} e^{- 4 x}

Schritte zur Lösung

y^{''} (x) + y = 3 e^{- 4 x}

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + \frac{3}{17} e^{- 4 x}

y = c_{1} cos (x) + c_{2} sin (x) + \frac{3}{17} e^{- 4 x}

\frac{d}{d x} (\frac{x sec ( - ( \frac{y}{x} ) + y )}{x})

\int (x^{2} - 1)^{3} d x

f (t) = 1 + t

\frac{d}{d x} (y (x))

y^{^{'}} + \frac{t y}{t ^{2}} = \frac{4}{t ^{2}}