2y^{''}-6y^'+5y=0

Lösung

2 y^{^{''}} - 6 y^{^{'}} + 5 y = 0

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

Schritte zur Lösung

2 y^{''} (t) - 6 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

\int cot (15 x) d x

\frac{\partial}{\partial s} (s^{2} + t^{2})

n = 0 \sum \infty n sin (n \cdot π)

in v erse l a pl a ce \frac{s + 1}{( s ( s ^{2} + s + 1 ) )}

\frac{\partial}{\partial x} (e^{c o s (x)})