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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 15 y = 0, y (0) = sin (k t)

y^{^{''}} - \frac{1}{c} y = 0

2 y^{^{''}} + 12 y = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = 4

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

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