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Beliebt Rechnen >

2y^{''}+25y=0

Lösung

2 y^{^{''}} + 25 y = 0

Lösung

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2})

Schritte zur Lösung

2 y^{''} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2})

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2})

Graph

Beliebte Beispiele

y^{''}+0.46y^'+128y=0

y^{^{''}} + 0.46 y^{^{'}} + 128 y = 0

2 y^{^{''}} - 4 y = 0

f^{''}(t)+2f^'(t)+5f(t)=0,f(0)=3,f^'(0)=-7

f^{^{''}} (t) + 2 f^{^{'}} (t) + 5 f (t) = 0, f (0) = 3, f^{^{'}} (0) = - 7

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

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

y^{''}-3y^'+2y=0,y(0)=-1,y^'(0)=2

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = - 1, y^{^{'}} (0) = 2