Using the method of variation of parameters find the general solution of the differential equation x^2 y'' - 3xy' + 3y = 12x^4 given that y1 = x and y2 = x^3 are solutions of the corresponding homogenous equation.
"x^{2}y^{''}-3xy^{'}+3y=12x^{4}"
"\\Rightarrow y^{''}-3\\frac{y^{'}}{x}+3\\frac{y}{x^{2}}=12x^{2}" "[ y^{''}+ay^{'}+by=R]"
Hence Wronskian (W)="y_{1}y_{2}^{'}-y_{2}y_{1}^{'}=x 3x^{2}-x^{3}1=2x^{3} \\neq 0"
Particular solution is "xf(x)+x^{3}g(x)" where "f(x)=-\\int\\frac{y_{2} R}{W}dx" "=-\\int \\frac{x^{3}12x^{2}}{2x^{3}} dx=-2x^{3}"
"g(x)=\\int \\frac{y_{1}R}{W} dx=\\int\\frac{x 12x^{2}}{x^{3}}dx=6x"
Therefore particular solution is "-2x^{3} x+x^{3}6x=4x^{4}"
The required solution is "y=c_{1}x+c_{2}x^{3}+4x^{4}" where "c_{1},c_{2}" are arbitrary constants.
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