The integrating factor
makes the differential equation
xydx + (2x2 + 3y2 - 20)dy = 0 exact.
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The equation is M(x,y)dx + N(x,y)dy =0 with M = xy , N= 2x^2 + 3y^2 - 20 . it is M_y ... The solution F(x,y) = C is obtained integrating the eq.
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equation is M(x,y)dx + N(x,y)dy =0 with M = xy , N= 2x^2 + 3y^2 - 20 .
it is M_y = x # N_x = 4x , but ( N_x - M_y)/M = 3/y , depends only on y , then
the integrating factor is I.F. = e^( 3lny ) = y^3 and produces the exact differential
dF(x,y) = Pdx + Qdy = 0 , with P(x,y) = xy^4 , Q(x,y) = 2x^2y^3 +3y^6 -20y^3 .
The solution F(x,y) = C is obtained integrating the eq. F_x = P wich gives
F(x,y) = (x^2y^4)/2 +G(y) ,Using the eq. F_y = Q leads to G’(y) = 3y^5 - 20y^3
i.e. G(y) = (1/2)y^6 - 5y^4 + C . Hence F(x,y) = ( y^4)( (x^2)/2 + y^2 -5 ) = C.
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