What is the solution for y^{''}-8y^'+7y=14 ?

The solution for y^{''}-8y^'+7y=14 is y=c_{1}e^{7t}+c_{2}e^t+2

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y^{''}-8y^'+7y=14

y^{^{''}} - 8 y^{^{'}} + 7 y = 14

y = c_{1} e^{7 t} + c_{2} e^{t} + 2

Solution steps

y^{''} (t) - 8 y^{'} (t) + 7 y = 14

Solve linear ODE:

y = c_{1} e^{7 t} + c_{2} e^{t} + 2

y = c_{1} e^{7 t} + c_{2} e^{t} + 2

y^{^{''}} - y^{^{'}} - 2 y = 3 sin (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 36 t + 6

y^{^{''}} + y^{^{'}} = 3 e^{\frac{x}{2}}

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 e^{5 t}

y^{^{''}} + 36 y = 8 cos (6 x) + 12 sin (6 x)

What is the solution for y^{''}-8y^'+7y=14 ?
The solution for y^{''}-8y^'+7y=14 is y=c_{1}e^{7t}+c_{2}e^t+2