0.125y^{''}+4y^'+20y=0

Lösung

0.125 y^{^{''}} + 4 y^{^{'}} + 20 y = 0

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

Schritte zur Lösung

0.125 y^{''} (t) + 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

\frac{\partial}{\partial y} (\frac{x}{1 + x ^{2} + y ^{2}})

\int_{0}^{1} \frac{9 x}{e ^{2 x}} d x

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\frac{d}{d x} (1 + \frac{3}{x})

\int \frac{1}{50 + t} d t