y^{''}+3y^'+2y=e-t,y(4)=1,y^'(4)=3

Lösung

y^{^{''}} + 3 y^{^{'}} + 2 y = e - t, y (4) = 1, y^{^{'}} (4) = 3

y = (\frac{- 4 e ^{9} + 20 e ^{8}}{4 e ^{4}} + 3 e^{4}) e^{- t} - \frac{- 2 e ^{9} + 23 e ^{8}}{4} e^{- 2 t} - \frac{t}{2} + \frac{2 e + 3}{4}

Schritte zur Lösung

y^{''} (t) + 3 y^{'} (t) + 2 y = e - t

Solve linear ODE:

y = (\frac{- 4 e ^{9} + 20 e ^{8}}{4 e ^{4}} + 3 e^{4}) e^{- t} - \frac{- 2 e ^{9} + 23 e ^{8}}{4} e^{- 2 t} - \frac{t}{2} + \frac{2 e + 3}{4}

y = (\frac{- 4 e ^{9} + 20 e ^{8}}{4 e ^{4}} + 3 e^{4}) e^{- t} - \frac{- 2 e ^{9} + 23 e ^{8}}{4} e^{- 2 t} - \frac{t}{2} + \frac{2 e + 3}{4}

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 x + 6

y^{^{''}} - 4 y = 28

y^{^{''}} + 4 y^{^{'}} + 16 y = 4 sin (6 t)

y^{^{''}} + 16 y = 16, y (0) = 1, y^{^{'}} (0) = 4