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

The solution for y^{''}-4y^'+3y=e^{(3x)} is y=c_{1}e^{3x}+c_{2}e^x+(e^{3x}x)/2

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 4 y = 7 e^{x}

y^{^{''}} + 25 y = 2 x, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} - 10 y^{^{'}} + 25 y = 2 e^{- 3 x}

y^{^{''}} - 4 y^{^{'}} - 12 y = t - 6

y^{^{''}} - 6 y^{^{'}} + 5 y = sin (4 x), y (0) = 1, y^{^{'}} (0) = - 4

What is the solution for y^{''}-4y^'+3y=e^{(3x)} ?
The solution for y^{''}-4y^'+3y=e^{(3x)} is y=c_{1}e^{3x}+c_{2}e^x+(e^{3x}x)/2