What is the solution for y^{''}-4y^'+3y=3+3x,y(0)=1,y^'(0)=3 ?

The solution for y^{''}-4y^'+3y=3+3x,y(0)=1,y^'(0)=3 is y=(3-4/3)e^{3x}-3e^x+x+7/3

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y^{''}-4y^'+3y=3+3x,y(0)=1,y^'(0)=3

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 + 3 x, y (0) = 1, y^{^{'}} (0) = 3

y = (3 - \frac{4}{3}) e^{3 x} - 3 e^{x} + x + \frac{7}{3}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 3 + 3 x

Solve linear ODE:

y = (3 - \frac{4}{3}) e^{3 x} - 3 e^{x} + x + \frac{7}{3}

y = (3 - \frac{4}{3}) e^{3 x} - 3 e^{x} + x + \frac{7}{3}

y^{^{'''}} - 3 y^{^{'}} - 2 y = e^{x} (x e^{x} + 1)

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

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

y^{^{'''}} - 21 y^{^{''}} + 1372 y = e^{14 x}

y^{^{''}} + 2 y^{^{'}} - 3 y = 4 e^{- 3 x}

What is the solution for y^{''}-4y^'+3y=3+3x,y(0)=1,y^'(0)=3 ?
The solution for y^{''}-4y^'+3y=3+3x,y(0)=1,y^'(0)=3 is y=(3-4/3)e^{3x}-3e^x+x+7/3