4y^{''}=y

Lösung

4 y^{^{''}} = y

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{2}}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{2}}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{- \frac{t}{2}}

y^{^{'}} = - 3 y

\frac{d}{d x} (ln (2 x - y))

\frac{d}{d x} (\frac{x ^{3}}{6})

y^{^{''}} + 9 y = 18

n = 1 \sum \infty \frac{15 n !}{n ^{n}}