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

Lösung

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

y = e^{\frac{5 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{\frac{5 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

y = e^{\frac{5 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

in v erse l a pl a ce \frac{e ^{- s} + s}{s ^{2} + s}

\frac{d}{d x} (\frac{2 ( 3 x ^{2} + 1 )}{( 1 - x ^{2} ) ^{3}})

\int_{- a}^{a} \int_{0}^{1} sin (y) e^{x^{2}} d x d y

f (x) = x 1 - x^{2}

\int_{1}^{8} \frac{1}{3 x + 1} d x