4y+y^{''}=0

Lösung

4 y + y^{^{''}} = 0

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

Schritte zur Lösung

4 y + y^{''} (t) = 0

Solve linear ODE:

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

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

d er i v a t i v e (\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 9 x^{3})

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

\frac{d ^{2} y}{d t ^{2}} + 10 \frac{d y}{d t} + 25 y = 0

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

t an g e n t f (x) = x - 4, (13, 3)