What is the solution for y^{''}+6y^'+5y=36e^x,y(0)=-2,y^'(0)=4 ?

The solution for y^{''}+6y^'+5y=36e^x,y(0)=-2,y^'(0)=4 is y=-6e^{-x}+e^{-5x}+3e^x

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y^{''}+6y^'+5y=36e^x,y(0)=-2,y^'(0)=4

y^{^{''}} + 6 y^{^{'}} + 5 y = 36 e^{x}, y (0) = - 2, y^{^{'}} (0) = 4

y = - 6 e^{- x} + e^{- 5 x} + 3 e^{x}

Solution steps

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

Solve linear ODE:

y = - 6 e^{- x} + e^{- 5 x} + 3 e^{x}

y = - 6 e^{- x} + e^{- 5 x} + 3 e^{x}

y^{^{''}} + y^{^{'}} - 12 y = \frac{e ^{x}}{x}

(6 D^{2} + 4 D + 2) y = x e^{- 2 x}

- 21 y + 4 y^{^{'}} + y^{^{''}} = t + 1

y^{^{''}} + 14 y^{^{'}} + 50 y = 6 e^{- 7 t} cos (t)

y, d (y) = 7 (2)^{y}

What is the solution for y^{''}+6y^'+5y=36e^x,y(0)=-2,y^'(0)=4 ?
The solution for y^{''}+6y^'+5y=36e^x,y(0)=-2,y^'(0)=4 is y=-6e^{-x}+e^{-5x}+3e^x