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Beliebt Rechnen >

5y^{''}+2y^'+7y=0,y(0)=2,y^'(0)=1

Lösung

5 y^{^{''}} + 2 y^{^{'}} + 7 y = 0, y (0) = 2, y^{^{'}} (0) = 1

Lösung

y = e^{- \frac{t}{5}} (2 cos (\frac{34 t}{5}) + \frac{7}{34} sin (\frac{34 t}{5}))

Schritte zur Lösung

5 y^{''} (t) + 2 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{- \frac{t}{5}} (2 cos (\frac{34 t}{5}) + \frac{7}{34} sin (\frac{34 t}{5}))

y = e^{- \frac{t}{5}} (2 cos (\frac{34 t}{5}) + \frac{7}{34} sin (\frac{34 t}{5}))

Graph

Beliebte Beispiele

y^{'''}+y^{''}-12y=0

y^{^{'''}} + y^{^{''}} - 12 y = 0

y^{''}-6y^'-10y=0

y^{^{''}} - 6 y^{^{'}} - 10 y = 0

y^{''}+(1+n^2)y=0

y^{^{''}} + (1 + n^{2}) y = 0

y^{''}+2y^'+2y=0,y(pi/2)=0,y^'(pi/2)=4

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 4

33y^{''''}+188y^{''}+64y=0

33 y^{^{''''}} + 188 y^{^{''}} + 64 y = 0