What is the solution for 3y^{''}+5y^'-y=10 ?

The solution for 3y^{''}+5y^'-y=10 is y=c_{1}e^{((-5+sqrt(37))t)/6}+c_{2}e^{((-5-sqrt(37))t)/6}-10

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3y^{''}+5y^'-y=10

3 y^{^{''}} + 5 y^{^{'}} - y = 10

y = c_{1} e^{\frac{( - 5 + 37 ) t}{6}} + c_{2} e^{\frac{( - 5 - 37 ) t}{6}} - 10

Solution steps

3 y^{''} (t) + 5 y^{'} (t) - y = 10

Solve linear ODE:

y = c_{1} e^{\frac{( - 5 + 37 ) t}{6}} + c_{2} e^{\frac{( - 5 - 37 ) t}{6}} - 10

y = c_{1} e^{\frac{( - 5 + 37 ) t}{6}} + c_{2} e^{\frac{( - 5 - 37 ) t}{6}} - 10

What is the solution for 3y^{''}+5y^'-y=10 ?
The solution for 3y^{''}+5y^'-y=10 is y=c_{1}e^{((-5+sqrt(37))t)/6}+c_{2}e^{((-5-sqrt(37))t)/6}-10