What is the solution for 2y^{''}+5y^'-y=9 ?

The solution for 2y^{''}+5y^'-y=9 is y=c_{1}e^{((-5+sqrt(33))t)/4}+c_{2}e^{((-5-sqrt(33))t)/4}-9

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2y^{''}+5y^'-y=9

2 y^{^{''}} + 5 y^{^{'}} - y = 9

y = c_{1} e^{\frac{( - 5 + 33 ) t}{4}} + c_{2} e^{\frac{( - 5 - 33 ) t}{4}} - 9

Solution steps

2 y^{''} (t) + 5 y^{'} (t) - y = 9

Solve linear ODE:

y = c_{1} e^{\frac{( - 5 + 33 ) t}{4}} + c_{2} e^{\frac{( - 5 - 33 ) t}{4}} - 9

y = c_{1} e^{\frac{( - 5 + 33 ) t}{4}} + c_{2} e^{\frac{( - 5 - 33 ) t}{4}} - 9

y^{^{''}} + 7 y^{^{'}} + 12 y = 340 cos (t)

y^{^{''}} + y = 2, + t y (0) = 1, y (0) = 1

y^{^{''}} + 25 y = 8 sec (5 x)

y^{^{''}} - y = (1 - e^{2 x})^{- \frac{1}{2}}

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 8 y = 8 e^{- 2 x}

What is the solution for 2y^{''}+5y^'-y=9 ?
The solution for 2y^{''}+5y^'-y=9 is y=c_{1}e^{((-5+sqrt(33))t)/4}+c_{2}e^{((-5-sqrt(33))t)/4}-9