What is the solution for 6y^{''}+4y^'-y=25 ?

The solution for 6y^{''}+4y^'-y=25 is y=c_{1}e^{((-2+sqrt(10))t)/6}+c_{2}e^{-((2+sqrt(10))t)/6}-25

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6y^{''}+4y^'-y=25

6 y^{^{''}} + 4 y^{^{'}} - y = 25

y = c_{1} e^{\frac{( - 2 + 10 ) t}{6}} + c_{2} e^{- \frac{( 2 + 10 ) t}{6}} - 25

Solution steps

6 y^{''} (t) + 4 y^{'} (t) - y = 25

Solve linear ODE:

y = c_{1} e^{\frac{( - 2 + 10 ) t}{6}} + c_{2} e^{- \frac{( 2 + 10 ) t}{6}} - 25

y = c_{1} e^{\frac{( - 2 + 10 ) t}{6}} + c_{2} e^{- \frac{( 2 + 10 ) t}{6}} - 25

10000 x^{^{''}} + 1000 x^{^{'}} + 22 x - 60000 = 0

y^{^{''}} + 3 y^{^{'}} + 6 \cdot x = 0

y^{^{''}} - 8 y^{^{'}} + 12 y = 2 x (e^{2 x} - e^{- 2 x})

4 q^{^{''}} + 8 q^{^{'}} + 80 q = 88 t

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 5

What is the solution for 6y^{''}+4y^'-y=25 ?
The solution for 6y^{''}+4y^'-y=25 is y=c_{1}e^{((-2+sqrt(10))t)/6}+c_{2}e^{-((2+sqrt(10))t)/6}-25