y^{''}+4y^'=t^2+2e^t,y(0)=0,y^'(0)=1

Lösung

y^{^{''}} + 4 y^{^{'}} = t^{2} + 2 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{91}{640} - \frac{2}{5} - \frac{91}{640} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{2}{5} e^{t}

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{91}{640} - \frac{2}{5} - \frac{91}{640} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{2}{5} e^{t}

y = \frac{91}{640} - \frac{2}{5} - \frac{91}{640} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{2}{5} e^{t}

y^{^{''}} + 3 y = 8 sin (t)

\frac{d ^{2} y}{d t ^{2}} - y = 10 e^{t}

y^{^{''}} - 0 \cdot y^{^{'}} - 2 y = - 2 t + 4 t^{2}

y^{^{''}} - y = - 20 t + 2

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = π e^{x} cos (x)