y^{''}+4y=4cos(2t),y(0)=0,y^'(0)=1

Lösung

y^{^{''}} + 4 y = 4 cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

y = t sin (2 t) + \frac{1}{2} sin (2 t)

Schritte zur Lösung

y^{''} (t) + 4 y = 4 cos (2 t)

Solve linear ODE:

y = t sin (2 t) + \frac{1}{2} sin (2 t)

y = t sin (2 t) + \frac{1}{2} sin (2 t)

y^{^{''}} + 4 y^{^{'}} = sin (x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{t}{e ^{t}} + \frac{e ^{t}}{t}

4 y^{^{''}} - 7 y^{^{'}} + 3 y = t e^{2 t} sin (- 2 t)

(D^{2} - 3 D + 2) y = e^{5 x}

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