Answer :
its simple homogeneous form....dy/dx=y-x/y+x....
then take y=vx....therefore ,
dy/dx=xdv/dx+v.....now solve step by step....easily the answer will come...
then take y=vx....therefore ,
dy/dx=xdv/dx+v.....now solve step by step....easily the answer will come...
y' = (y - x)/(y + x)
Dividing both the numerator and denominator of the fraction on the right side by x:
y' = [(y/x) - 1]/[(y/x) + 1]
Let y/x = u
==> y = ux ==> y' = u'x + u
The equation becomes
u'x + u = (u - 1)/(u + 1)
u'x = (u - 1)/(u + 1) - u
u'x = -(u² + 1)/(u + 1)
u' = -(u² + 1)/(u + 1)x
[(u + 1)/(u² + 1)]u' = -1/x
[(u + 1)/(u² + 1)]du = -dx/x
udu/(u² + 1) + du/(u² + 1) = -dx/x
2udu/(u² + 1) + 2du/(u² + 1) = -2dx/x
Intergrating each side:
ln(u² + 1) + 2arctan(u) = -ln(x²) + C
ln(u² + 1) + ln(x²) + 2arctan(u) = C
ln(u²x² + x²) + 2arctan(u) = C
ln(y² + x²) + 2arctan(y/x) = C. . . (answer)
Dividing both the numerator and denominator of the fraction on the right side by x:
y' = [(y/x) - 1]/[(y/x) + 1]
Let y/x = u
==> y = ux ==> y' = u'x + u
The equation becomes
u'x + u = (u - 1)/(u + 1)
u'x = (u - 1)/(u + 1) - u
u'x = -(u² + 1)/(u + 1)
u' = -(u² + 1)/(u + 1)x
[(u + 1)/(u² + 1)]u' = -1/x
[(u + 1)/(u² + 1)]du = -dx/x
udu/(u² + 1) + du/(u² + 1) = -dx/x
2udu/(u² + 1) + 2du/(u² + 1) = -2dx/x
Intergrating each side:
ln(u² + 1) + 2arctan(u) = -ln(x²) + C
ln(u² + 1) + ln(x²) + 2arctan(u) = C
ln(u²x² + x²) + 2arctan(u) = C
ln(y² + x²) + 2arctan(y/x) = C. . . (answer)