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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} + 3 D + 2) y = 0

y^{^{''}} - \frac{10}{3} y^{^{'}} + a y = 0

x^{^{''}} + a \cdot x^{^{'}} = 0

y^{^{'''}} + y^{^{''}} - 5 y^{^{'}} + 3 y = 0

y^{^{'''''}} - 2 y^{^{''''}} + 4 y^{^{'''}} - 8 y^{^{''}} = 0

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