y^{''}+4y=t^2+8e^t,y(0)=0,y^'(0)=7

Lösung

y^{^{''}} + 4 y = t^{2} + 8 e^{t}, y (0) = 0, y^{^{'}} (0) = 7

y = - \frac{59}{40} cos (2 t) + \frac{27}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{8}{5} e^{t}

Schritte zur Lösung

y^{''} (t) + 4 y = t^{2} + 8 e^{t}

Solve linear ODE:

y = - \frac{59}{40} cos (2 t) + \frac{27}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{8}{5} e^{t}

y = - \frac{59}{40} cos (2 t) + \frac{27}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{8}{5} e^{t}

y^{^{''}} - 4 y = e^{x} cos (x), y (0) = 0.9, y^{^{'}} (0) = - 1.5

\frac{d ^{2} y}{d x ^{2}} + y = 3 sin (2 x)

y^{^{''}} - 4 y = e^{x} cos (x), y (0) = 0.9, y^{^{'}} (0) = - 1.1

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

y^{^{''}} - y^{^{'}} - 6 y = 28 cos (3 x)