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

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

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

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

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

Solution steps

2 y^{''} (t) - 2 y^{'} (t) + 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}))

\int \frac{1}{t} d t

\frac{d}{d x} (- \frac{1}{( x + 4 ) ^{2}})

f (x) = 4 (x^{2} + 4 x)^{3}

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

\frac{d}{d x} (5 + tan^{2} (x))

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