t^2y^{''}+ty^'+25y=0

Lösung

t^{2} y^{^{''}} + t y^{^{'}} + 25 y = 0

y = c_{1} cos (5 ln (t)) + c_{2} sin (5 ln (t))

Schritte zur Lösung

t^{2} y^{''} (t) + t y^{'} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} cos (5 ln (t)) + c_{2} sin (5 ln (t))

y = c_{1} cos (5 ln (t)) + c_{2} sin (5 ln (t))

\frac{d y}{d x} = 3 x

x \to - 1 lim (- x^{2} + 1)

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 3

\int_{0}^{1} \frac{2}{t ^{2} + 1} d t

d er i v a t i v e f (x) = 2 π r^{2} + \frac{1000}{r}