What is the solution for 3y^{''}-4y^'+7y=0 ?

The solution for 3y^{''}-4y^'+7y=0 is y=e^{(2t)/3}(c_{1}cos((sqrt(17)t)/3)+c_{2}sin((sqrt(17)t)/3))

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3y^{''}-4y^'+7y=0

3 y^{^{''}} - 4 y^{^{'}} + 7 y = 0

y = e^{\frac{2 t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

Solution steps

3 y^{''} (t) - 4 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{\frac{2 t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

y = e^{\frac{2 t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = e^{t}

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y^{^{''}} + 16 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 12

y^{^{''}} + 3 y^{^{'}} + 7 y = 0

y^{^{''}} + y^{^{'}} = y

What is the solution for 3y^{''}-4y^'+7y=0 ?
The solution for 3y^{''}-4y^'+7y=0 is y=e^{(2t)/3}(c_{1}cos((sqrt(17)t)/3)+c_{2}sin((sqrt(17)t)/3))