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if a curve y=f(x) passes through the point (1,-1)satisfies the differential equation y(1+xy) dx= x dy then f(-1/2)
write it asydx + xy^2dx= xdyor (ydx – xdy)/y^2 = – xdxnow note that (ydx – xdy)/y^2 = d(x/y)so d(x/y)= – xdxintegrate both sidesx/y= – x^2/2 + C(1,-1)satisfies the differential equation so – 1= – ½ + Cor C= – ½ so, y=f(x)= – 2x/(1+x^2)so f(-1/2)= 4/5= 0.8kindly approve :))
Dear student y/x (1+xy) = dy/dx y = vx = y/x = v dy/dx = v+xdv/dx v(1+vx^2) = v+xdv/dx v^2x^2 = xdv/dx Integrate this x^2/2 = -1/v + c x^2/2 = -x/y - 1/2 f(-1/2) , put x = -1/2 1/8 = 1/2y - 1/2 y = 4/5
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