Juan Hewlett

2021-12-21

What is the differential equation of the orthogonal trajectories of the family of curves ${x}^{2}-3xy-2{y}^{2}=C$?

autormtak0w

Expert

Introduction:
This straightforward question based on differential equations could be resolved by applying the fundamental understanding of the subject as
Solution:
Given: ${x}^{2}-3xy-2{y}^{2}=C$
Now differentiating w.r.t. x we get

Therefore $2x-3y-{y}^{\prime }\left(3x+4y\right)=0$ is the required differential equation.

Annie Gonzalez

Expert

$\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{dy}{dx}$
Given, ${x}^{2}-3xy-2{y}^{2}=c$
Differentiating both sides
$\frac{d}{dx}\left({x}^{2}-3xy-2{y}^{2}\right)=\frac{d}{dx}\left(c\right)$
$⇒\frac{d}{dx}\left({x}^{2}\right)-\frac{d}{dx}\left(3xy\right)-\frac{d}{dx}\left(2{y}^{2}\right)=0$
$⇒2x-3\left(x\frac{dy}{dx}+y\frac{dx}{dx}\right)-2×2y\frac{dy}{dx}=0$
$⇒2x-3x\frac{dy}{dx}-3y-4y\frac{dy}{dx}=0$
$⇒\left(2x-3y\right)=3x\frac{dy}{dx}+4y\frac{dy}{dx}$
$⇒\left(2x-3y\right)=\left(3x+4y\right)\frac{dy}{dx}$ ...(1)
Replacing $\frac{dy}{dx}=-\frac{dx}{dy}$ in equation (i), we get
$\left(2x-3y\right)=\left(3x+4y\right)\cdot \left(-\frac{dx}{dy}\right)$
$⇒\left(2x-3y\right)dy=-\left(3x+4y\right)dx$

RizerMix

Expert

Given family of Curve is ${x}^{2}-3xy-2{y}^{2}=C$
Since ${x}^{2}-3xy-2{y}^{2}=c$
Diff. both side with respect to x, we get

Hence $\left(3x+4y\right)dx+\left(2x-3y\right)dy=0$ is the differential equation of the orthogonal trajectories of family of given curve.

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