y^{''}+16y=cos(4t),y(0)=3,y^'(0)=2

Lösung

y^{^{''}} + 16 y = cos (4 t), y (0) = 3, y^{^{'}} (0) = 2

y = 3 cos (4 t) + \frac{1}{2} sin (4 t) + \frac{t sin ( 4 t )}{8}

Schritte zur Lösung

y^{''} (t) + 16 y = cos (4 t)

Solve linear ODE:

y = 3 cos (4 t) + \frac{1}{2} sin (4 t) + \frac{t sin ( 4 t )}{8}

y = 3 cos (4 t) + \frac{1}{2} sin (4 t) + \frac{t sin ( 4 t )}{8}

3 \frac{d ^{2} y}{d x ^{2}} + 20 \frac{d y}{d x} + 25 y = cos (2 x)

(D^{2} + 4 D + 5) y = e^{(- x)} + 15 x

y^{^{''}} + y = 2 cos (x), y^{^{'}} (π) = 0, y (0) = 0

(D^{2} + 1) y = cos^{2} (x)

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 9 y = x e^{- 3 x}