What is the solution for 2y^{''}-4y^'+3y=0 ?

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

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2y^{''}-4y^'+3y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y^{^{'}} (0) = 0

\frac{1}{8} x^{^{''}} + 16 x = 0

y^{^{'''}} - 27 y^{^{''}} + 243 y^{^{'}} - 729 y = 0

y^{^{''}} + 4 y = 0, y^{^{'}} (3 π) = 4, y (0) = 4

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