y^{''}+4y=e^t,y(0)=0,y^'(0)=0

Lösung

y^{^{''}} + 4 y = e^{t}, y (0) = 0, y^{^{'}} (0) = 0

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

Schritte zur Lösung

y^{''} (t) + 4 y = e^{t}

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = x^{2} e^{x}

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} - 3 y = e^{t}

y^{^{''}} + 8 y^{^{'}} + 17 y = 6 e^{- 40} cos (t)

y^{^{''}} + \frac{3}{2} y^{^{'}} = x + e^{2 x}

\frac{d ^{4} y}{d x ^{4}} + 6 \frac{d y}{d x} = - 5 x^{2}