Notes On Homogeneous Differential Equations - CBSE Class 12 Maths
Consider two functions F(x,y) = (2x2y2)/(x2+y2) G(x,y) = y2/(x2 - xy) Substitute kx in place of x, and ky in place of y where k is a constant. F(kx,ky) = (2k2x2k2y2)/(k2x2+k2y2) , G(kx,ky) = k2y2/(k2x2 - kx.ky) F(kx,ky) = k4/k2 . (2x2y2)/(x2+y2) = k2 (2x2y2)/(x2+y2) ⇒ F(kx,ky) = k2F(x,y) G(kx,ky) = k2/k2 . y2/(x2 - xy) = k0. y2/(x2 - xy) ⇒ G(kx,ky) = k0G(x,y) This type of functions is known as homogeneous functions. Here, the function F(x, y) is of degree two, and the function G (x, y) is of degree zero. Any homogeneous function can be expressed as F(x,y) = xng(y/x) F(x,y) = ynh(x/y) n = Degree of homogeneous equation We consider the first function F(x,y) = (2x2y2)/(x2+y2) = (2x2[x2.(y/x)2])/(x2.(1+ y2/x2)) = (2x4.(y/x)2])/(x2.(1+ y2/x2)) = (x2. 2(y/x)2])/(1+ (y/x)2) = x2F1(y/x) This is a homogeneous function of degree two. Similarly, G(x,y) = y2/(x2-xy) = y2/(y2[(x/y)2 - (x/y)])  = y2-2. 1/[(x/y)2 - (x/y)] = y0 G1(x/y) Here, zero is the degree of the function G (x, y). A differential equation of the form dy/dx = F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero. Consider a homogeneous differential equation of degree zero. dy/dx = x0h(y/x) ................... (i) Substitute y = vx. dy/dx = v + x. dv/dx equation (i) can be written as v + x . dv/dx= x0h(v) Or v + x dv/dx = h(v) x dv/dx = h(v) - v ⇒ 1/(h(v) - v) dv = 1/x dx ⇒ ∫ 1/(h(v) - v) dv = ∫ 1/x dx Replace v = y/x For the homogeneous function in the differential equation is of the form dy/dx = y0h(x/y) Substitute x = vy in the equation.

#### Summary

Consider two functions F(x,y) = (2x2y2)/(x2+y2) G(x,y) = y2/(x2 - xy) Substitute kx in place of x, and ky in place of y where k is a constant. F(kx,ky) = (2k2x2k2y2)/(k2x2+k2y2) , G(kx,ky) = k2y2/(k2x2 - kx.ky) F(kx,ky) = k4/k2 . (2x2y2)/(x2+y2) = k2 (2x2y2)/(x2+y2) ⇒ F(kx,ky) = k2F(x,y) G(kx,ky) = k2/k2 . y2/(x2 - xy) = k0. y2/(x2 - xy) ⇒ G(kx,ky) = k0G(x,y) This type of functions is known as homogeneous functions. Here, the function F(x, y) is of degree two, and the function G (x, y) is of degree zero. Any homogeneous function can be expressed as F(x,y) = xng(y/x) F(x,y) = ynh(x/y) n = Degree of homogeneous equation We consider the first function F(x,y) = (2x2y2)/(x2+y2) = (2x2[x2.(y/x)2])/(x2.(1+ y2/x2)) = (2x4.(y/x)2])/(x2.(1+ y2/x2)) = (x2. 2(y/x)2])/(1+ (y/x)2) = x2F1(y/x) This is a homogeneous function of degree two. Similarly, G(x,y) = y2/(x2-xy) = y2/(y2[(x/y)2 - (x/y)])  = y2-2. 1/[(x/y)2 - (x/y)] = y0 G1(x/y) Here, zero is the degree of the function G (x, y). A differential equation of the form dy/dx = F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero. Consider a homogeneous differential equation of degree zero. dy/dx = x0h(y/x) ................... (i) Substitute y = vx. dy/dx = v + x. dv/dx equation (i) can be written as v + x . dv/dx= x0h(v) Or v + x dv/dx = h(v) x dv/dx = h(v) - v ⇒ 1/(h(v) - v) dv = 1/x dx ⇒ ∫ 1/(h(v) - v) dv = ∫ 1/x dx Replace v = y/x For the homogeneous function in the differential equation is of the form dy/dx = y0h(x/y) Substitute x = vy in the equation.

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