y^{''}-4y^'+4y=2t^2+4te^{2t}+tsin(2t)

Lösung

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{2} + 4 t e^{2 t} + t sin (2 t)

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{2}}{2} + t + \frac{3}{4} + \frac{2 e ^{2 t} t ^{3}}{3} + \frac{t cos ( 2 t )}{8} + \frac{1}{16} cos (2 t) - \frac{1}{16} sin (2 t)

Schritte zur Lösung

y^{''} (t) - 4 y^{'} (t) + 4 y = 2 t^{2} + 4 t e^{2 t} + t sin (2 t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{2}}{2} + t + \frac{3}{4} + \frac{2 e ^{2 t} t ^{3}}{3} + \frac{t cos ( 2 t )}{8} + \frac{1}{16} cos (2 t) - \frac{1}{16} sin (2 t)

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{2}}{2} + t + \frac{3}{4} + \frac{2 e ^{2 t} t ^{3}}{3} + \frac{t cos ( 2 t )}{8} + \frac{1}{16} cos (2 t) - \frac{1}{16} sin (2 t)

y^{^{''}} - 12 y^{^{'}} + 36 y = (x + 1) e^{x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- t}}{t + 1}

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

4 y^{^{''}} + 16 y = sin (x) + 2 cos (x)

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