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

The solution for 2y^{''}+3y^'-y=0 is y=c_{1}e^{((-3+sqrt(17))t)/4}+c_{2}e^{((-3-sqrt(17))t)/4}

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

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

y = c_{1} e^{\frac{( - 3 + 17 ) t}{4}} + c_{2} e^{\frac{( - 3 - 17 ) t}{4}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 17 ) t}{4}} + c_{2} e^{\frac{( - 3 - 17 ) t}{4}}

y = c_{1} e^{\frac{( - 3 + 17 ) t}{4}} + c_{2} e^{\frac{( - 3 - 17 ) t}{4}}

What is the solution for 2y^{''}+3y^'-y=0 ?
The solution for 2y^{''}+3y^'-y=0 is y=c_{1}e^{((-3+sqrt(17))t)/4}+c_{2}e^{((-3-sqrt(17))t)/4}