What is the solution for y^{''}=7y^'+4y ?

The solution for y^{''}=7y^'+4y is y=c_{1}e^{((7+sqrt(65))t)/2}+c_{2}e^{((7-sqrt(65))t)/2}

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y^{''}=7y^'+4y

y^{^{''}} = 7 y^{^{'}} + 4 y

y = c_{1} e^{\frac{( 7 + 65 ) t}{2}} + c_{2} e^{\frac{( 7 - 65 ) t}{2}}

Solution steps

y^{''} (t) = 7 y^{'} (t) + 4 y

Solve linear ODE:

y = c_{1} e^{\frac{( 7 + 65 ) t}{2}} + c_{2} e^{\frac{( 7 - 65 ) t}{2}}

y = c_{1} e^{\frac{( 7 + 65 ) t}{2}} + c_{2} e^{\frac{( 7 - 65 ) t}{2}}

y^{^{''}} + 0.05 y^{^{'}} + 25 y = 0

2 \cdot y^{^{''}} + 4 \cdot y^{^{'}} + 8 y = 0

5 y^{^{''}} + 18 y^{^{'}} + 9 y = 0, y (0) = 17.4

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (1) = 3

y^{^{''}} + 96 y = 0

What is the solution for y^{''}=7y^'+4y ?
The solution for y^{''}=7y^'+4y is y=c_{1}e^{((7+sqrt(65))t)/2}+c_{2}e^{((7-sqrt(65))t)/2}