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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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