What is the solution for y^{''}-5y^'+2y=0 ?

The solution for y^{''}-5y^'+2y=0 is y=c_{1}e^{((5+sqrt(17))t)/2}+c_{2}e^{((5-sqrt(17))t)/2}

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y^{''}-5y^'+2y=0

y^{^{''}} - 5 y^{^{'}} + 2 y = 0

y = c_{1} e^{\frac{( 5 + 17 ) t}{2}} + c_{2} e^{\frac{( 5 - 17 ) t}{2}}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{( 5 + 17 ) t}{2}} + c_{2} e^{\frac{( 5 - 17 ) t}{2}}

y = c_{1} e^{\frac{( 5 + 17 ) t}{2}} + c_{2} e^{\frac{( 5 - 17 ) t}{2}}

What is the solution for y^{''}-5y^'+2y=0 ?
The solution for y^{''}-5y^'+2y=0 is y=c_{1}e^{((5+sqrt(17))t)/2}+c_{2}e^{((5-sqrt(17))t)/2}