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

The solution for y^{''}-4y^'+3y=x,y(0)=1,y^'(0)=0 is y=(-1+5/9)e^{3x}+e^x+x/3+4/9

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

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

y = (- 1 + \frac{5}{9}) e^{3 x} + e^{x} + \frac{x}{3} + \frac{4}{9}

Solution steps

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

Solve linear ODE:

y = (- 1 + \frac{5}{9}) e^{3 x} + e^{x} + \frac{x}{3} + \frac{4}{9}

y = (- 1 + \frac{5}{9}) e^{3 x} + e^{x} + \frac{x}{3} + \frac{4}{9}

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What is the solution for y^{''}-4y^'+3y=x,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-4y^'+3y=x,y(0)=1,y^'(0)=0 is y=(-1+5/9)e^{3x}+e^x+x/3+4/9