y^{'''}+3y^{''}+2y^'=(x^2+4x+8)e^2x

Lösung

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = (x^{2} + 4 x + 8) e^{2} x

y = c_{1} + c_{2} e^{- x} + c_{3} e^{- 2 x} + \frac{e ^{2} x ^{4}}{8} - \frac{e ^{2} x ^{3}}{12} + \frac{13 e ^{2} x ^{2}}{8} - \frac{37 e ^{2} x}{8}

Schritte zur Lösung

y^{'''} (x) + 3 y^{''} (x) + 2 y^{'} (x) = (x^{2} + 4 x + 8) e^{2} x

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + c_{3} e^{- 2 x} + \frac{e ^{2} x ^{4}}{8} - \frac{e ^{2} x ^{3}}{12} + \frac{13 e ^{2} x ^{2}}{8} - \frac{37 e ^{2} x}{8}

y = c_{1} + c_{2} e^{- x} + c_{3} e^{- 2 x} + \frac{e ^{2} x ^{4}}{8} - \frac{e ^{2} x ^{3}}{12} + \frac{13 e ^{2} x ^{2}}{8} - \frac{37 e ^{2} x}{8}

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 5 e x

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

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

y^{^{''}} + y = 3 t^{2}, y (0) = 0, y^{^{'}} (0) = 0

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