y^{''}-5y^'+7y=2

Lösung

y^{^{''}} - 5 y^{^{'}} + 7 y = 2

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + \frac{2}{7}

Schritte zur Lösung

y^{''} (t) - 5 y^{'} (t) + 7 y = 2

Solve linear ODE:

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + \frac{2}{7}

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})) + \frac{2}{7}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 2 y = 3 e^{x} sec (x)

y^{^{''}} - 25 y = 6 sin (x)

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

y^{^{''}} + 2 y^{^{'}} + 6 y = e^{- 2 t}, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + 5 y = 1, y (0) = 0, y^{^{'}} (0) = 0