Answer to Question #189657 in Differential Equations for pedma

Question #189657

(x^2-y^2 -zy)p +(x^2-y^2-zx)q =z(x-y)


1
Expert's answer
2021-05-11T14:15:54-0400

By lagrange's auxilary equation - "\\dfrac{dx}{x^{2}-y^{2}-zy}=\\dfrac{dy}{x^{2}-y^{2}-zx}=\\dfrac{dz}{z(x-y)}"


now in first two ratio , deviding numerator and denominator by x and y respectively ,


"= \\dfrac{xdx}{x^{3}-xy^{2}-xzy}=\\dfrac {ydy}{x^{2}y-y^{3}-zxy}" = "\\dfrac {{dz}\/{z}}{{x-y}}"


"=" "\\dfrac{xdx-ydy}{x^{3}+y^{3}-xy(x+y)}" "=\\dfrac{dz\/z}{x-y}"


"=\\dfrac{xdx-ydy}{(x+y)(x-y)^2}=\\dfrac{dz\/z}{x-y}"



"=\\dfrac{xdx-ydy}{x^{2}-y^{2}}=\\dfrac{dz}{z}"


"=\\dfrac{d(x^{2}-y^{2})}{2(x^{2}-y^{2})}" "=\\dfrac{dz}{z}" , now integerating both sides we get ,


"=z^{2}=x^{2}-y^{2}+c_1" which is required solution.







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