4y^{''}+16y=17sec^2(4t)

Lösung

4 y^{^{''}} + 16 y = 17 sec^{2} (4 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{17 sec ^{2} ( 4 t ) sin ( 2 t )}{8} d t + sin (2 t) \cdot \int \frac{17 sec ^{2} ( 4 t ) cos ( 2 t )}{8} d t

Schritte zur Lösung

4 y^{''} (t) + 16 y = 17 sec^{2} (4 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{17 sec ^{2} ( 4 t ) sin ( 2 t )}{8} d t + sin (2 t) \cdot \int \frac{17 sec ^{2} ( 4 t ) cos ( 2 t )}{8} d t

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{17 sec ^{2} ( 4 t ) sin ( 2 t )}{8} d t + sin (2 t) \cdot \int \frac{17 sec ^{2} ( 4 t ) cos ( 2 t )}{8} d t

- y^{^{''}} + y = e^{1 \cdot x}, y (0) = 0, y (1) = 0

y^{^{''}} + 3 y^{^{'}} - 4 y = cos (3 t)

y^{^{''}} - 9 y = 24

\frac{d ^{2} y}{d t ^{2}} - 6 \frac{d y}{d t} + 9 y = t

y^{^{''}} - y^{^{'}} - 2 y = 3 cos (2 x)