y^{''}-y=1,y^'(0)=1

Lösung

y^{^{''}} - y = 1, y^{^{'}} (0) = 1

y = (1 + c_{2}) e^{t} + c_{2} e^{- t} - 1

Schritte zur Lösung

y^{''} (t) - y = 1

Solve linear ODE:

y = (1 + c_{2}) e^{t} + c_{2} e^{- t} - 1

y = (1 + c_{2}) e^{t} + c_{2} e^{- t} - 1

y^{^{''}} + 2 y^{^{'}} - 3 y = 6 x + 5

5 y^{^{''}} + 4 y^{^{'}} + 4 y = 35 e^{- 4 x}

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

y^{^{''}} + 3 y^{^{'}} + 4 = 0

y^{^{''}} + 9 y = 3 sec^{3} (3 x)