solve the DE .. x(dy/dx) - (1+x)y = xy^2 Solution x ( dy(x))/( dx)+ (-x-1) y(x)

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x ( dy(x))/( dx)+ (-x-1) y(x) = x y(x)^2 Divide both sides by -x y(x)^2: ==> -(( dy(x))/( dx))/y(x)^2+(1/x+1)/(y(x)) = -1 Let v(x) = 1/(y(x)) ==> ( dv(x))/( dx) = -(( dy(x))/( dx))/y(x)^2: ==>( dv(x))/( dx)+(1/x+1) v(x) = -1 Let mu(x) = exp( integral (1/x+1) dx) = e^x x. Multiply both sides by mu(x): ==>(e^x x) ( dv(x))/( dx)+(e^x x (1/x+1)) v(x) = -e^x x Substitute e^x x (1/x+1) = ( d)/( dx)(e^x x): ( e^x x) ( dv(x))/( dx)+( d)/( dx)(e^x x) v(x) = -e^x x ( d)/( dx)((e^x x) v(x)) = -e^x x integral ( d)/( dx)((e^x x) v(x)) dx = integral -e^x x dx (e^x x) v(x) = -e^x (x-1)+c , mu(x) = e^x x: v(x) = (-x+c e^(-x)+1)/x y(x) = 1/(v(x)) = (e^x* x)/(-e^x (x-1)+ c)

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