The particular integral of \(\frac{{{d^3}y}}{{d{x^3}}} - 3\frac{{{d^2}y}}{{d{x^2}}} + 4y = {e^{3x}}\) is
1
\(x\frac{{{e^{3x}}}}{9} + \frac{{2{x^2}}}{9}\)
2
\(x\frac{{{e^{3x}}}}{9} + \frac{{2{x^3}}}{9}\)
3
\(x\frac{{{e^{3x}}}}{9} + \frac{{2{(x^2+x^3)}}}{9}\)
4
\(\frac{{{e^{3x}}}}{4}\)