y^{''}+4y= t/2-(tcos(2t))/2

Lösung

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t}{8} - \frac{t ^{2} sin ( 2 t )}{16} - \frac{t cos ( 2 t )}{32}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t}{8} - \frac{t ^{2} sin ( 2 t )}{16} - \frac{t cos ( 2 t )}{32}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t}{8} - \frac{t ^{2} sin ( 2 t )}{16} - \frac{t cos ( 2 t )}{32}

y^{^{''}} - 7 y^{^{'}} - 8 y = 8 x^{2} - 2 x

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

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