y^{''}-12y^'+36=0

Lösung

y^{^{''}} - 12 y^{^{'}} + 36 = 0

y = c_{1} e^{12 t} + c_{2} + 3 t

Schritte zur Lösung

y^{''} (t) - 12 y^{'} (t) + 36 = 0

Solve linear ODE:

y = c_{1} e^{12 t} + c_{2} + 3 t

y = c_{1} e^{12 t} + c_{2} + 3 t

y^{^{''}} + 4 y = 2 cos (x) + 3 sin (x), y (0) = 1, y^{^{'}} (0) = 0

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

y^{^{''}} + 9 y = sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} + 5 y = x sin (x)

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