Homogeneous function- Question -1 A Differential equation of the form, dy/dx = f(x,y)/?(x,y) Where f(x,y) and ?(x,y) are homogeneous function of same degree.
Such equations can be solved by putting , y= v.x Dy/dx =dy/dx= v-xdy/dx Putting the value of y and then solve by variable - separable method Example - x2dy +y(x+y0 dx=0 Solution- Dy/dx = dy/dx = -y(x+y) /x2 Let y = vx ==( dy/dx = v+xdv/dx Now , v+xdv/dx = - x(vx) + v2x2/x2 = -(v2+v) ===( xdv/dx = -( v2-2v) Integrating ,?dv/v2+2v = -?dx/x ==(?dv/v(v+2) = -?dx/x Or , ?1/2 (1/v -1/v+2) dv = -?dx/x ½ lnv - ½ ln ( v+2) = -lnx+C OR , ½ log v/v+2 = -log x +C ==( ½ log ( y/x)/(y/x) +2 = - logx+C ===( log y/2x+y = ln 1/x2 +c ==( y/2x+y = C'/x2 ==( x2.y = K' ( 2x+y) Homogeneous function-
Question -1 A Differential equation of the form, dy/dx = f(x,y)/?(x,y) Where f(x,y) and ?(x,y) are homogeneous function of same degree. Such equations can be solved by putting , y= v.x Dy/dx =dy/dx= v-xdy/dx Putting the value of y and then solve by variable - separable method Example - x2dy +y(x+y0 dx=0
Solution- Dy/dx = dy/dx = -y(x+y) /x2 Let y = vx ==( dy/dx = v+xdv/dx Now , v+xdv/dx = - x(vx) + v2x2/x2 = -(v2+v) ===( xdv/dx = -( v2-2v) Integrating ,?dv/v2+2v = -?dx/x ==(?dv/v(v+2) = -?dx/x Or , ?1/2 (1/v -1/v+2) dv = -?dx/x ½ lnv - ½ ln ( v+2) = -lnx+C OR , ½ log v/v+2 = -log x +C ==( ½ log ( y/x)/(y/x) +2 = - logx+C ===( log y/2x+y = ln 1/x2 +c ==( y/2x+y = C'/x2 ==( x2.y = K' ( 2x+y)