4y^{''}+4y^'+9y=0

Lösung

4 y^{^{''}} + 4 y^{^{'}} + 9 y = 0

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

Schritte zur Lösung

4 y^{''} (t) + 4 y^{'} (t) + 9 y = 0

Solve linear ODE:

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

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

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\frac{d}{d x} (\frac{( x ^{2} - 5 ) ^{3}}{x ^{2} + 4})

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\frac{\partial}{\partial x} (3 z e^{x y})