y^{''}-0.02y=0,y^'(0)=2,y(0)=0

Lösung

y^{^{''}} - 0.02 y = 0, y^{^{'}} (0) = 2, y (0) = 0

y = 52 e^{\frac{t}{5 2}} - 52 e^{- \frac{t}{5 2}}

Schritte zur Lösung

y^{''} (t) - 0.02 y = 0

Solve linear ODE:

y = 52 e^{\frac{t}{5 2}} - 52 e^{- \frac{t}{5 2}}

y = 52 e^{\frac{t}{5 2}} - 52 e^{- \frac{t}{5 2}}

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