What is the solution for 4y^{''}-y^'-6y=0 ?

The solution for 4y^{''}-y^'-6y=0 is y=c_{1}e^{((1+sqrt(97))t)/8}+c_{2}e^{((1-sqrt(97))t)/8}

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4y^{''}-y^'-6y=0

4 y^{^{''}} - y^{^{'}} - 6 y = 0

y = c_{1} e^{\frac{( 1 + 97 ) t}{8}} + c_{2} e^{\frac{( 1 - 97 ) t}{8}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( 1 + 97 ) t}{8}} + c_{2} e^{\frac{( 1 - 97 ) t}{8}}

y = c_{1} e^{\frac{( 1 + 97 ) t}{8}} + c_{2} e^{\frac{( 1 - 97 ) t}{8}}

y^{^{''}} - \frac{2}{3} y^{^{'}} + \frac{y}{9} = 0, y (0) = 17, y^{^{'}} (0) = b

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} + 10 y + 16 y = 0

8 \frac{d ^{2} y}{d t ^{2}} + 12 \frac{d y}{d t} + 5 y = 0

\frac{d ^{2} y}{d t ^{2}} - 18 \frac{d y}{d t} + 81 y = 0

What is the solution for 4y^{''}-y^'-6y=0 ?
The solution for 4y^{''}-y^'-6y=0 is y=c_{1}e^{((1+sqrt(97))t)/8}+c_{2}e^{((1-sqrt(97))t)/8}