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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{'''}} + 6 y^{^{''}} - 27 y^{^{'}} = 0

\frac{d ^{2} y}{d t ^{2}} - 7 \frac{d y}{d t} = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (ln (2)) = 3, y (0) = 33

y^{^{'''}} - 4 y^{^{''}} - 45 y^{^{'}} = 0

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