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Beliebt Rechnen >

25y^{''}-40y^'+16y=0,y(0)=3,y^'(0)=-2

Lösung

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

Lösung

y = 3 e^{\frac{4 t}{5}} - \frac{22 e ^{\frac{4 t}{5}} t}{5}

Schritte zur Lösung

25 y^{''} (t) - 40 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = 3 e^{\frac{4 t}{5}} - \frac{22 e ^{\frac{4 t}{5}} t}{5}

y = 3 e^{\frac{4 t}{5}} - \frac{22 e ^{\frac{4 t}{5}} t}{5}

Graph

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