The auxiliary equation of the given equation is
"(m-1)^2=0"
"m_{1,2}=1"
Then the complementary function of the given equation is
When "f(x, y)=V," where "V" is function of "x" and "y," then partial integral:
Then partial integral
"=\\dfrac{1}{(D-D')^2 }(12xy)=\\dfrac{12}{D^2 }(1-\\dfrac{D'}{D})^{-2}(xy)="
"=\\dfrac{12}{D^2 }(1+\\dfrac{2D'}{D}+\\dfrac{3D'^2}{D^2}+...)(xy)="
"=\\dfrac{12}{D^2 }(xy+\\dfrac{2}{D}(x)+0+...)="
"=\\dfrac{12}{D^2 }(xy)+\\dfrac{24}{D^3 }(x)=12y(\\dfrac{x^3}{6 })+24(\\dfrac{x^4}{24 })="
Then the general solution of the given equation is "z=u+p," such that
where "\\varphi_1, \\varphi_2" are arbitrary functions.Â
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