y^{''}+4y^'+6.25y=3.125(x+1)^2

Lösung

y^{^{''}} + 4 y^{^{'}} + 6.25 y = 3.125 (x + 1)^{2}

y = e^{- 2 x} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + \frac{x ^{2}}{2} + \frac{9 x}{25} + \frac{137}{1250}

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) + 6.25 y = 3.125 (x + 1)^{2}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + \frac{x ^{2}}{2} + \frac{9 x}{25} + \frac{137}{1250}

y = e^{- 2 x} (c_{1} cos (\frac{3 x}{2}) + c_{2} sin (\frac{3 x}{2})) + \frac{x ^{2}}{2} + \frac{9 x}{25} + \frac{137}{1250}

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

\frac{d}{d x} (ln^{2} (1 - 2 x))

d er i v a t i v e f (x) = \frac{x}{ln ( x )}

y^{^{''}} + 6 y^{^{'}} + 19 y = 0

y^{^{''}} = 0