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Lecture 3 [MATH 129]

 Differential Equation

Formation of Differential Equation

[Continues]


    Question: Form the differential equation of the curve represented by y22ay+x2=a2. where a is an arbitrary constant.

    Solution: The given equation is,

  y22ay+x2=a2  

Differentiating both sides of (1.1) with respect to x we get,

  2ydydx2adydx+2x=0ydydx+x=adydxy+xdxdy=a  

Substituting this value of a in (1.1) we have, 

  y22h(y+xdxdy)+x2=(y+xdxdy)2y22y22xydxdy+x2=y2+2xydxdy+x2(dxdy)22y2+4xydxdy+x2(dxdy)2x2=02y2(dydx)2+4xydydx+x2x2(dydx)2=0  




    

    Question: Form the differential equation of all straight lines passing through the origin.

    Solution: We know, the equation of all straight lines passing through the origin is,

  y=mx  

Differentiating (2.1).

  dydx=m  

eliminating m between (2.1) and (2.2)

i.e by putting    m=dydx   in  y=mx

we get,  y=xdydx  

Which is the required differential equation. 






    Question: Form the differential equation corresponding to \(y = ae^{2x} + be^{-3x}+ ce^x \), where a,b,c are arbitrary constantx.

    Solution: The given equation is, 

&nbsp y=ae2x+be3x+cex  

Differentiating both sides of (3.1) with respect o x, we get 

  dydx=2ae2x3be3x+cex  

Subtracting (3.1) from (3.2) we get, 

  dydxy=ae2x4be3x  

Differentiating (3.3) we get,

  d2ydx2dydx=3ae2x+12be3x  

Adding 3 times of (3.3) to (3.4) we have,

 (d2ydx2dydx)+d(dydxy)=(2ae2x+12be3x)+3(ae2x4be3x)d2ydx2+2dydx3y=5ae2x  

Differentiating both sides of (3.5) with respect to x we get,

  d3ydx3+2d2ydx2dydx=10ae2x  

From (3.5) and (3.6) eliminating a we get,

&nbsp d3ydx3+2d2ydx23dydx=2(d2ydx2+2dydx3y)  d3ydx37dydx+6y=0  





    Question: Find the differential equation of the family of curves xy=Aex+Bex+x2 .

    Solution: Given equation is, 

  xy=Aex+Bex+x2

differentiating with respect to x we get,

  xdydx+y=AexBex+2x  

Differentiating again we get,

  xd2ydx2+dydx+dydx=Aex+Bex+2xd2ydx2+2dydx=xyx2+2  

Thus the two variables A and B have been eliminated and we get the differential equation of the 2nd order as,

  xd2ydx2+2dydx=xyx2+2  







    Question: Find the differential equation of the family of curves y=ex(Acosx+Bsinx), where A and B are arbitrary constants.

    Solution: The given equation is,

  y=ex(Acosx+Bsinx)  

Differentiating with respect to x, we get

  dydx=ex(Asinx+Bcosx)+ex(Acosx+Bsinx)dydx=ex(Asinx+Bcosx)+y  

Again Differentiating both sides with respect to x, we get

  d2ydx2=ex(AcosxBsinx)+ex(Asinx+Bcosx)+dydx=ex(Acosx+Bsinx)+(dydxy)+dydxy+(dydxy)+dydx d2ydx22dydx+2y=0  





    Question: Find the differential equation of the family y=cx3.

    Solution: The given equation is,

  y=cx3c=yx3  

Now 

  y=cx3dydx=3cx2dydx=3yx3x2dydx=3yxxdydx=3yxdydx3y=0  

which is the required differential equation.





    Homeworks:

1. Show that the differential equation of the family of the circle touches the x-axis at the origin is  (x2y2) dy2xy dx=0  

2. Find the Differential Equation of the family of curves  y=ex(Acosx+Bsinx) where A and B are arbitrary constants.

3. Discuss the degree and order of the differential equations with examples.

4. Find the differential equation of all circles of radius a.

5. Form a differential equation from  y=Acosax+Bsinax , where A and B arbitrary constants and a is a fixed constant.

6. Find the differential equation of all circles which pass through the origin and whose centers are on the x-axis.

7. briefly discuss differential equation with its degree and order. Form an ordinary differential equation from  y=ex(acosx+bsinx) , where a and b parameter.  











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