y^{''}-4y^'+13y=sin(x)-e^{3x}+5

Lösung

y^{^{''}} - 4 y^{^{'}} + 13 y = sin (x) - e^{3 x} + 5

y = e^{2 x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{3}{40} sin (x) + \frac{1}{40} cos (x) - \frac{1}{10} e^{3 x} + \frac{5}{13}

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 13 y = sin (x) - e^{3 x} + 5

Solve linear ODE:

y = e^{2 x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{3}{40} sin (x) + \frac{1}{40} cos (x) - \frac{1}{10} e^{3 x} + \frac{5}{13}

y = e^{2 x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{3}{40} sin (x) + \frac{1}{40} cos (x) - \frac{1}{10} e^{3 x} + \frac{5}{13}

t an g e n t (x^{3} - 16 x)^{8}

\frac{\partial}{\partial y} (2 x^{4} y + 4)

\frac{d}{d x} (sec (0))

\int (4 - x^{2})^{\frac{1}{2}} d x

\frac{d y}{d t} = - \frac{3}{4 t}