What is the solution for 49y^{''}+140y^'+136=0 ?

The solution for 49y^{''}+140y^'+136=0 is y=c_{1}+c_{2}e^{-(20t)/7}-(34t)/(35)

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49y^{''}+140y^'+136=0

49 y^{^{''}} + 140 y^{^{'}} + 136 = 0

y = c_{1} + c_{2} e^{- \frac{20 t}{7}} - \frac{34 t}{35}

Solution steps

49 y^{''} (t) + 140 y^{'} (t) + 136 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{20 t}{7}} - \frac{34 t}{35}

y = c_{1} + c_{2} e^{- \frac{20 t}{7}} - \frac{34 t}{35}

x^{^{''}} + \frac{1}{98} x = 8 cos (6 t)

33 y^{^{''}} - 22 y^{^{'}} - 3 y = 8 - t

y^{^{''}} + y = 2 cos (x), y (0) = 1

y^{^{''}} + 2 y^{^{'}} + 5 y = 10

y^{^{'''}} + y^{^{'}} = t^{2} + 1

What is the solution for 49y^{''}+140y^'+136=0 ?
The solution for 49y^{''}+140y^'+136=0 is y=c_{1}+c_{2}e^{-(20t)/7}-(34t)/(35)