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

Lösung

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

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)

Schritte zur Lösung

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

Solve linear ODE:

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)

y^{^{''}} + 4 y^{^{'}} - 21 y = 5 e^{2 x} + 4

y^{^{''}} + 4 y = 1 - t^{3}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = sin (2 x)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 1 + x - 7 e^{- 2 x}