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

The solution for y^{''}-3y^'+y=0 is y=c_{1}e^{((3+sqrt(5))t)/2}+c_{2}e^{((3-sqrt(5))t)/2}

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

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

y = c_{1} e^{\frac{( 3 + 5 ) t}{2}} + c_{2} e^{\frac{( 3 - 5 ) t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( 3 + 5 ) t}{2}} + c_{2} e^{\frac{( 3 - 5 ) t}{2}}

y = c_{1} e^{\frac{( 3 + 5 ) t}{2}} + c_{2} e^{\frac{( 3 - 5 ) t}{2}}

y = 3 x^{3} - 3 x

\frac{d y}{d x} = 5 e^{x}

\frac{d}{d x} (ln (1 + cos (x)))

\int_{0}^{1} x arcsin (x^{2}) d x

\frac{d}{d x} (1200 x^{2} - 0.1 x)

What is the solution for y^{''}-3y^'+y=0 ?
The solution for y^{''}-3y^'+y=0 is y=c_{1}e^{((3+sqrt(5))t)/2}+c_{2}e^{((3-sqrt(5))t)/2}