TYPE–5. HOMOGEMEOUS EQUATIONS : A differential equation of the form (dy/dx) =(f(x,y))/( Æ(x,y))
where f (x, y) & Æ (x, y) are homogeneous functions of x & y, and of the same degree, is called HOMOGENEOUS. This equation may also be reduced to the form (dy/dx) =g (x/y) & is solved by putting y = vx so that the dependent variable y changed to another variable v, where v is some unknown function, the differential equation is transformed to an equation with variables separable.
Consider . (dy/dx) + (y(x,y)/x2) = 0 IMPORTANT NOTE :
(a) The function f (x, y) is said to be a homogeneous function of degree n if for any real number t(± 0) , we have f (tx, ty) = tn f (x, y). For e. g. f (x, y) = ax2/3 + hx1/3 . y1/3 + by2/3 is homogeneous function of degree 2/3.
(b) A differential equation of the form (dy/dx) = f (x,y) is homogeneous if f (x, y) is a homogeneous function of degree zero i. e. f (tx, ty) = t0 f (x, y) = f (x, y).
The function f does not depend on x & y separately but only on the their ratio y/x or x/y. EXAMPLES ON HOMOGENEOUS :
(i) (dy/dx) = (y2 2xy –x2) / (y2 +2xy-x2) given | y |x=1 = –1 [Ans x + y = 0]
[Hint: ∫ (v2+2v-1)/((1+v)(1+v)2) dv= ∫ (2v(v+1)-(1-v2)) / ((1+v)(1+v2)) dv =- ∫ (dx/d)
(ii) {x(dy/dx) –y} tan -1 (y/x) = x given | y |x=1 = 0
[Solution : x(dy/dx) –(y/x) ) .tan-1 (y/x) =x
put y = vx Þ (dy/dx) = v+ x (dv/dx)
{v+ x (dv/dx) –v} tan-1 v=1
x (dv/dx) tan-1 v=1
∫ tan-1 v dv = dx/x
v tan-1 v- ∫ (v/(1+v2)) dv = Inx + C
where x = 1 ; y = 0 Þ v = 0 Þ C = 0
(y/x) tan-1 (y/x) = In Ö{1+ (y2/x2) } + Inx =Inx Ö{1+(y2/x2)} =inq(x2+y2)
Ö(x2 +y2) = e(y/x)tan-1(y/x)]
Some times a homogeneous differential equation is given in the form of a quadratic equation in (dy/dx) (i.e. degree 2, consider the example)
(iii) y(dy/dx)2 +2x(dy/dx) –y =0 |y|x=0 =Ö 5 ;
[Solution : (dy/dx) = (-x ±Ö(4x2+ 4y2)/(2y)) = (dy/dx) (-x±Ö(x2 +y2))/(y)) Which is homogeneous
Taking +ve sign
|||ly with –ve sign,y2 =5 -2 Ö5 x ]
(iv) Misleading appearance of a D. E. containing terms like (y/x) & (x/y), however not homogeneous and this can not be solved by substituting y = vx.
[Solution : Isolating
However note that the D.E.
{x cos (y/x) + y sin (y/x) } ydx +{ xcos (y/x) –y sin( y/x) xdy =0 is homogeneous and can be solved conviniently with y = vx
simplifies to
(v) Storey Problems The area of the figure bounded by a curve, the x-axis and tow ordinates, one of which is constant, the other variable is equal to the ratio of the cube of the variable ordinate to the variable abscissa. Find the curve if it passes through (1,0).
[Ans. (2y – x2)3 = cx2]
[Solution: c ∫ ydx =(y3/ x)
y= (x.3y2 y1 -y3 /x2)
x2 3xyy1 -y2 (y¹0)
(dy/dx) = (x2 +y2 /3xy) which is Homogeneous.
(vi) Find the equation of the curve intersecting with the x-axis at the point x = 1 and possesses the property that the length of the subnormal at any point of the curve is equal to the A.M. of the co-ordinates of this point. [Ans.(y–x)2 (x+2y) = 1 ]
(vii) Find the curve such that the ordinate of any of its points is the proportional mean between the abscissa and the sum of the abscissa and subnormal at the point.
[Ans. Y2 =(x4 +c4 /2x2 or y2 +2x2 Inx =cx2 ]
[Hint: The differential equation y2 = x { x ± y ( dy/dx) }]
(viii) Find the curve such that the angle, formed with the x-axis by the tangent to the curve at any of its points, is twice the angle formed by the polar radius of the point of tangency with the x-axis. Interpret the curve.
Given tan2q = (dy/dx)
where tan q =(y/x)
or
[Ans: x2 + y2 – 2cy = 0 ]
Note that the curve is a circle with centre on y-axis and touching the x-axis at the origin.
(ix) Find the curve such that the ratio of the subnormal at any point to the sum of its abscissa and ordinates is equal to the ratio of the ordinate of this point to its abscissa.
[Solution:
with + ve sign Solution is y = x ln |cx |
with –ve sign Solution is x2 + 2xy = c ]
(x) Find curve for which the sum of the normal and subnormal is proportional to the abscissa.
[ Solution: Refer to the figure of (ix)
Ö(m2y2+y2)+ Ö(my) =kx
|y|Ö(1+m2) kx -|my|
squaring y2(1+m2) = (k2x2) + m2y2 - 2kx |my|
2kx |my| = k2x2 - y2
(dy/dx) = ± (k2x2 -y2)/(2kxy]
