What is the solution for y^{''}+6y^'+9y=3x+5,y(0)=1,y^'(0)=-1 ?

The solution for y^{''}+6y^'+9y=3x+5,y(0)=1,y^'(0)=-1 is y= 2/3 e^{-3x}+(2e^{-3x}x)/3+x/3+1/3

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y^{''}+6y^'+9y=3x+5,y(0)=1,y^'(0)=-1

y^{^{''}} + 6 y^{^{'}} + 9 y = 3 x + 5, y (0) = 1, y^{^{'}} (0) = - 1

y = \frac{2}{3} e^{- 3 x} + \frac{2 e ^{- 3 x} x}{3} + \frac{x}{3} + \frac{1}{3}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 9 y = 3 x + 5

Solve linear ODE:

y = \frac{2}{3} e^{- 3 x} + \frac{2 e ^{- 3 x} x}{3} + \frac{x}{3} + \frac{1}{3}

y = \frac{2}{3} e^{- 3 x} + \frac{2 e ^{- 3 x} x}{3} + \frac{x}{3} + \frac{1}{3}

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

y^{^{''}} + 2 y = - 3

y^{^{''}} + 8 y = 2 t^{2} - 2, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} + 25 y = 64 cos (3 x), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} - 2 y = - 2 x^{3} - 3 x^{2} + 8 x + 1

What is the solution for y^{''}+6y^'+9y=3x+5,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+6y^'+9y=3x+5,y(0)=1,y^'(0)=-1 is y= 2/3 e^{-3x}+(2e^{-3x}x)/3+x/3+1/3