What is the solution for y^{''}+6y^'+5y=36e^x,y(0)=-2 ?

The solution for y^{''}+6y^'+5y=36e^x,y(0)=-2 is y=(-c_{2}-5)e^{-x}+c_{2}e^{-5x}+3e^x

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y^{''}+6y^'+5y=36e^x,y(0)=-2

y^{^{''}} + 6 y^{^{'}} + 5 y = 36 e^{x}, y (0) = - 2

y = (- c_{2} - 5) e^{- x} + c_{2} e^{- 5 x} + 3 e^{x}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 5 y = 36 e^{x}

Solve linear ODE:

y = (- c_{2} - 5) e^{- x} + c_{2} e^{- 5 x} + 3 e^{x}

y = (- c_{2} - 5) e^{- x} + c_{2} e^{- 5 x} + 3 e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y (0) = - 1

y^{^{''}} + y = e^{t}

y^{^{''}} - 2 y^{^{'}} + 2 y = \frac{e ^{x}}{cos ( x )}

y^{^{'''}} + 9 y^{^{'}} = - 18 t

y^{^{''}} + 16 y = 5 sec (4 x)

What is the solution for y^{''}+6y^'+5y=36e^x,y(0)=-2 ?
The solution for y^{''}+6y^'+5y=36e^x,y(0)=-2 is y=(-c_{2}-5)e^{-x}+c_{2}e^{-5x}+3e^x