Homogeneous Equations and Solutions

Edited with the trial version of Foxit Advanced PDF Editor To remove this notice, visit: www.foxitsoftware.com/shopping Homogeneous Function f (x, y) and n is homogeneous function then f ( x, y) = nf (x, y) represents the degree of the homogeneous function. If Example For the function f (x, y) = x2 + y2 then f ( x, y) =( x)2 +( y)2 = 2x2 + 2y2 = 2(x2 + y2)= 2 f (x, y) So, the function f (x, y) is homogeneous with degree 2. Example For the function f (x, y) = x + y2 then f ( x, y) = x+( y)2 (x, y) = x3+ xy + x f (x, (Non-homogeneous) f (x, y) = x2+ y2+ 5 f (Non-homogeneous) = x+ 2 y2 = (x+ y2) (Non-homogeneous) (Non-homogeneous) y) = cos(xy) f (x, y) = cos(x2 y2) 1 So, the function f (x, y) is not homogeneous. Example

Edited with the trial version of Foxit Advanced PDF Editor To remove this notice, visit: www.foxitsoftware.com/shopping x f (x, y) = cos f (x, y) = cos (Homogeneous) x2 y (Non-homogeneous) y Homogeneous Equations and N The differential equation M(x, y)dx + N(x, y)dy is homogeneous if M are homogeneous functions of the same degree. Example 1) (x2 + y2)dx + xydy = 0 This is homogeneous because M and N are both homogeneous with degree 2. 2) (x3 + y3)dx + xydy = 0 This is not homogeneous because M is homogeneous with degree 3 while N is homogeneous with degree 2. 3) xdx+(x2 + y)dy = 0 Solution of Homogeneous Equations This is not homogeneous because N is not homogeneous. Ahomogeneous first order differential equation can be put in the form y x dy = F dx This equation can be changed into separable equation with the substitutions 2

Edited with the trial version of Foxit Advanced PDF Editor To remove this notice, visit: www.foxitsoftware.com/shopping v =y y = vx v + xdv= F(v) dx dy= v+ xdv dx x dx Then becomes which can be rearranged algebraically to give dx+ x dv = 0 v F(v) with the variables now separated, the equation can now be solved by integrating with respect to x and v. We can then return to x and y by substituting v = y/ x. x2 + y2 dy dx = 2xy Example that satisfies the condition y(1) =1. Find the solution of the differential equation Solution dy 1+(y/ x)2 dx v F(v) = v +1 1+v2 = dy dx Dividing the numerator and denominator of the right-hand side by x2gives 2(y/ x) = = F(v) 2v 2v2 +1+v2 3v2 +1 = dx+2vdv = 0 x 3v2 +1 +v 2 = 2v 2v 2v dx+ x dv = 0 v F(v) The solution of this equation can be written as 3

Edited with the trial version of Foxit Advanced PDF Editor To remove this notice, visit: www.foxitsoftware.com/shopping lnx +1ln(1+3v2) = C 3 2vdv dx = C 3ln x +ln(1+3v2) = 3C + 3v2 +1 x ln x3+ln(1+3v2) = 3C 3 2 elnx +ln(1+3v ) = e3C x3(1+3v2) =C 3 2 2 y elnx eln(1+3v ) = e3C x 1+3 = C x2 3 x3+3xy2=C The condition is that when x =1 then y =1 and the constant C can be found (1)3+3(1)(1)2=C C = 4 The final solution is x3+3xy2= 4. Reducible to Homogeneous If the differential equation has the form a2x +b2y +c2 dx dy=a1x+b1y +c1 then z = a1x+b1y a1=b1 a2 Case 1: if b2 a1x+b1y +c1 = 0 Case 2: then intersect the two if a2x+b2y +c2 = 0 to find the intersection point (h,k) and let lines and a2 a1 b1 b2 x = X + h dx = dX , and y =Y +k dy = dY 4

Edited with the trial version of Foxit Advanced PDF Editor To remove this notice, visit: www.foxitsoftware.com/shopping Example 4x+6y +5 dx dy = Solve the differential equation 3y +2x+4 Solution dy=2x+3y +4 dx 4x+6y +5 2 4 1 2 a1 a2 = = a1 = 2, a2 = 4 = = 6 3 2 1 b2 b1 Case 1 So b1 =3, z = 2x+3y =1 dz 2 dz=3z +12+4z +10 dx 2z +5 7z +22 dz 7 7 7 2z 9 ln(7z +22) x =C 7 49 2(2x +3y) 9 ln(7(2x +3y)+22) x =C 7 49 b2 = 6 dz= 2+3dy dx Let a1=b1 a2 dx dx =3z +12+2 2z +5 2z +5 dz = dx 7z +22 dz dx =C b2 3 dx 2=z +4 3 dx 2z +5 7 dz dx =C 1 3 dx dx =7z +22 2 9 2z +5 2z +5 dz dx =C 9 1 7z +22 7 2 7 7z +22 5

Edited with the trial version of Foxit Advanced PDF Editor To remove this notice, visit: www.foxitsoftware.com/shopping Example Solve the differential equation (2x + y 3)dy = (x +2y 3)dx dy=x+2y 3 dx 2x+ y 3 Solution 1 a1 = 2 a2 b1 b2 a1 =1, a2 = 2 = 2 Case 2 So b1 = 2, x 2x +2x m 2x b2 =1 + 2y + y m y 3y y =1 3 3 = 0 = 0 (1) (2) .. .. a1 b1 a2 = 0 = 0 = 0 6 3 3 +4y b2 Substituting into (2), we get 2x+1 3= 0 The intersection point (h,k) = (1,1). Let x = X +1 y =Y +1 dY=(X +1)+2(Y +1) 3 dX 2(X +1)+(Y +1) 3 =X +1+ 2Y +2 3 2X + 2+Y +1 3 x =1 dx = dX dy = dY dY=X + 2Y dX 2X +Y 6