4y^{''}-y^'+6=0

Lösung

4 y^{^{''}} - y^{^{'}} + 6 = 0

y = c_{1} e^{\frac{t}{4}} + c_{2} + 6 t

Schritte zur Lösung

4 y^{''} (t) - y^{'} (t) + 6 = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{4}} + c_{2} + 6 t

y = c_{1} e^{\frac{t}{4}} + c_{2} + 6 t

y^{^{''}} + 4 y = 2 e^{t}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 2 \cdot y = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{( 16 e ^{- 3 x} )}{x ^{2} + 1}

y^{^{''}} - 2 y^{^{'}} - 24 y = 72 e^{2 t}

y^{^{''}} + 49 y = t^{2} cos (7 t) + t e^{(- 7 t)}