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)

Question

Aditya Gupta · Answer

write it as ydx + xy^2dx= xdy or (ydx – xdy)/y^2 = – xdx now note that (ydx – xdy)/y^2 = d(x/y) so d(x/y)= – xdx integrate both sides x/y= – x^2/2 + C (1,-1)satisfies the differential equation so – 1= – ½ + C or C= – ½ so, y=f(x)= – 2x/(1+x^2) so f(-1/2)= 4/5= 0.8 kindly approve :))

Vikas TU · Answer

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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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)

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)

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