What is the solution for 2y^{''}-6y^'+y=0 ?

The solution for 2y^{''}-6y^'+y=0 is y=c_{1}e^{((3+sqrt(7))t)/2}+c_{2}e^{((3-sqrt(7))t)/2}

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2y^{''}-6y^'+y=0

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

y = c_{1} e^{\frac{( 3 + 7 ) t}{2}} + c_{2} e^{\frac{( 3 - 7 ) t}{2}}

Solution steps

2 y^{''} (t) - 6 y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 3 + 7 ) t}{2}} + c_{2} e^{\frac{( 3 - 7 ) t}{2}}

y = c_{1} e^{\frac{( 3 + 7 ) t}{2}} + c_{2} e^{\frac{( 3 - 7 ) t}{2}}

\int 5 sec^{2} (x) d x

6 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

\int \frac{2 x + 1}{x ^{2} + 2 x + 10} d x

x \to 2 lim (x^{2} - 3 x + 1)

x \to 2 lim ([1 + f (x)]^{2})

What is the solution for 2y^{''}-6y^'+y=0 ?
The solution for 2y^{''}-6y^'+y=0 is y=c_{1}e^{((3+sqrt(7))t)/2}+c_{2}e^{((3-sqrt(7))t)/2}