What is the solution for y^{''}+10y^'+25=0 ?

The solution for y^{''}+10y^'+25=0 is y=c_{1}+c_{2}e^{-10t}-(5t)/2

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y^{''}+10y^'+25=0

y^{^{''}} + 10 y^{^{'}} + 25 = 0

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

Solution steps

y^{''} (t) + 10 y^{'} (t) + 25 = 0

Solve linear ODE:

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

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

\int (\frac{100}{25 + ( 5 y ) ^{2}} + e^{4}) d y

\int \frac{x ^{2}}{( x - 1 )} d x

f (x) = - 2 x^{2}, at x = - 2

4 x - 16

What is the solution for y^{''}+10y^'+25=0 ?
The solution for y^{''}+10y^'+25=0 is y=c_{1}+c_{2}e^{-10t}-(5t)/2