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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{( x - 1 )}{x + 1})

x \to \infty lim (\frac{x ^{2} - 9}{2 x - 6})

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x}

\frac{d}{d x} (4 x^{2} + 6 x)

\frac{\partial}{\partial s} (s^{6} + t^{6})