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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 8 y = sin (x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 12 cos (2 x) + 3 sin (2 x)

y^{^{''}} + 4 y = 2 x, y (0) = 1, y^{^{'}} (0) = 2

\frac{d ^{2} u}{d x ^{2}} - u = 6

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = e^{x}