y^{''}-2y^'=(4t+9)sin(2t)

Lösung

y^{^{''}} - 2 y^{^{'}} = (4 t + 9) sin (2 t)

y = c_{1} e^{2 t} + c_{2} + \frac{t cos ( 2 t )}{2} - \frac{t sin ( 2 t )}{2} + \frac{7}{8} cos (2 t) - \frac{13}{8} sin (2 t)

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) = (4 t + 9) sin (2 t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} + \frac{t cos ( 2 t )}{2} - \frac{t sin ( 2 t )}{2} + \frac{7}{8} cos (2 t) - \frac{13}{8} sin (2 t)

y = c_{1} e^{2 t} + c_{2} + \frac{t cos ( 2 t )}{2} - \frac{t sin ( 2 t )}{2} + \frac{7}{8} cos (2 t) - \frac{13}{8} sin (2 t)

\frac{d}{d x} (\frac{3}{2} sin (2 x))

\frac{d}{d x} (2 sin (x))

\int x csc^{2} (4 x) d x

\int_{- 1}^{1} 0.09375 (4 - x^{2}) d x

\int 24 x (6 - x^{2})^{\frac{1}{3}} d x