y^{'''}+2y^{''}+2y^'=x^2-sin(x)

Lösung

y^{^{'''}} + 2 y^{^{''}} + 2 y^{^{'}} = x^{2} - sin (x)

y = c_{1} + e^{- x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{x ^{3}}{6} + \frac{- x ^{2} + x}{2} + \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

Schritte zur Lösung

y^{'''} (x) + 2 y^{''} (x) + 2 y^{'} (x) = x^{2} - sin (x)

Solve linear ODE:

y = c_{1} + e^{- x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{x ^{3}}{6} + \frac{- x ^{2} + x}{2} + \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

y = c_{1} + e^{- x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{x ^{3}}{6} + \frac{- x ^{2} + x}{2} + \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

\frac{d ^{2} y}{d x ^{2}} + 4 y = cos (\frac{9}{4} x)

\frac{d ^{2} y}{d x ^{2}} + a y^{^{'}} = b

y^{^{''}} + y = 2 sin (2) t

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

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