Question

Solve  differiantial equation (y+x) dy/dx=y-x

 

Answer 1

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...

Answer 2

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)