What is the differential equation of the orthogonal trajectories of the family of curves x2−3xy−2y2=C?

Juan Hewlett

Juan Hewlett

Answered

2021-12-21

What is the differential equation of the orthogonal trajectories of the family of curves x23xy2y2=C?

Answer & Explanation

autormtak0w

autormtak0w

Expert

2021-12-22Added 31 answers

Introduction:
This straightforward question based on differential equations could be resolved by applying the fundamental understanding of the subject as
Solution:
Given: x23xy2y2=C
Now differentiating w.r.t. x we get
2x3(y+xy)2(2y)y=0 as (xy)=xy+xy  and  (C)=0
2x3y3xy4yy=0  or  2x3yy(3x+4y)=0
Therefore 2x3yy(3x+4y)=0 is the required differential equation.

Annie Gonzalez

Annie Gonzalez

Expert

2021-12-23Added 41 answers

ddx(uv)=udvdx+vdydx
Given, x23xy2y2=c
Differentiating both sides
ddx(x23xy2y2)=ddx(c)
ddx(x2)ddx(3xy)ddx(2y2)=0
2x3(xdydx+ydxdx)2×2ydydx=0
2x3xdydx3y4ydydx=0
(2x3y)=3xdydx+4ydydx
(2x3y)=(3x+4y)dydx ...(1)
Replacing dydx=dxdy in equation (i), we get
(2x3y)=(3x+4y)(dxdy)
(2x3y)dy=(3x+4y)dx
RizerMix

RizerMix

Expert

2021-12-29Added 437 answers

Given family of Curve is x23xy2y2=C
Since x23xy2y2=c
Diff. both side with respect to x, we get
ddx[x23xy2y2]=ddx[c]2x3(xdydx+y)4ydydx=02x3xdydx3y4ydydx=0dydx(3x4y)=3y2xput dydx=dxdydxdy(3x4y)=3y2x(3x+4y)dx=(3y2x)dy(3x+4y)dx+(2x3y)dy=0
Hence (3x+4y)dx+(2x3y)dy=0 is the differential equation of the orthogonal trajectories of family of given curve.

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