What is the solution for y^{''}+5y^'+6=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+5y^'+6=0,y(0)=1,y^'(0)=2 is y=1+16/25-16/25 e^{-5t}-(6t)/5

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y^{''}+5y^'+6=0,y(0)=1,y^'(0)=2

y^{^{''}} + 5 y^{^{'}} + 6 = 0, y (0) = 1, y^{^{'}} (0) = 2

y = 1 + \frac{16}{25} - \frac{16}{25} e^{- 5 t} - \frac{6 t}{5}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 6 = 0

Solve linear ODE:

y = 1 + \frac{16}{25} - \frac{16}{25} e^{- 5 t} - \frac{6 t}{5}

y = 1 + \frac{16}{25} - \frac{16}{25} e^{- 5 t} - \frac{6 t}{5}

y^{^{''}} + 11 y = 5 e^{3 x}

y^{^{''}} + 6 y^{^{'}} + 10 y = 180 sin (2 x)

y^{^{''}} + y^{^{'}} - 72 y = - 220 t + 144 t^{2}

y^{^{''}} - 10 y^{^{'}} + 25 y = 8 e^{5 x}

y^{^{''}} - y^{^{'}} - 2 y = 5 x + 5

What is the solution for y^{''}+5y^'+6=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+5y^'+6=0,y(0)=1,y^'(0)=2 is y=1+16/25-16/25 e^{-5t}-(6t)/5