Lecture 2 [MATH 129]

Differential Equation

Formation of Differential Equation


    Question: Find the differential equation that describes the family of circles passing through the origin.

    Solution: The general form of the equation of the circle is 

  (xh)2+(yk)2=(h2+k2)2x22xh+h2+y22yk+k2=h2+k2x22xh+y22ky=0(xh)2+(yk)2=(h2+k2)2x22xh+h2+y22yk+k2=h2+k2x22xh+y22ky=0(1.1)  

Using differentiation twice we get,

  2x2h+2yy2ky=0xh+yyky=0and 10+yy+yyky=01+yy+(y)2ky=0  

From the equation (1.1) we get,

  2xh=x2+y22kyh=(x2+y22ky)2x  

From equation (1.2) and (1.4) we get,

  xx2+y22ky2x+yyky=02x2x2y2+2ky+2xyy2xky=02k(yxy)=x2+y22xyy          k=x2+y22xyy2(yxy)          k=x2y2+2xyy2(xyy)  

Again from (1.3) and (1.5) we get,

  1+yy+(y)2x2y2+2xyy2(xyy)y=02[1+(y)2](xyy)+2yy(xyy)(x2y2+2xyy)y=02[1+(y)2](xyy)+2xyyy2y2yx2y+y22xyyy=0  

Which is the required differential equation.






    Question: Find the differential equation of the family  y=2ce2x1+ce2x  
    Solution: 
Given Equation, 
  y=2ce2x1+ce2xdydx=(1+ce2x) ddx(2ce2x)2ce2xddx(1+ce2x)(1+ce2x)2=(1+ce2x) 4ce2x2ce2x 2ce2x(1+ce2x)2=4ce2x+4c2e2x e2x4c2e2x e2x(1+ce2x)2=4ce2x(1+ce2x)2=(2ce2x1+ce2x)2 e2xcdydx=y2e2xc  
Again, 
           y=2ce2x1+ce2xy(1+ce2x)=2ce2xy+cye2x=2ce2xce2x(y2)=y       ce2x=y2y         c=ye2x2y  
From (2.1) and (2.2) we get, 
  dydx=y2e2x1e2xy2ydydx=y(2y)  
Which is the required differential equation.
    
















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