Question

Differential equation of a family of curve is given by ${\left(y{y}^{\prime }\right)}^2+2xy{y}^{\prime }{y}^2=0$, what is the differential equation of the family of its orthogonal trajectory?

• A. ${\left(y{y}^{\prime }\right)}^2+2xy{y}^{\prime }{y}^2=0$
• B. $y^{\prime }+2xy=0$
• C. ${\left(y{y}^{\prime }\right)}^2-2xy{y}^{\prime }{y}^2=0$
• D. $y^{\prime }-2xy=0$

Answer 1

The correct option is A ${\left(y{y}^{\prime }\right)}^2+2xy{y}^{\prime }{y}^2=0$

We have, ${\left(y{y}^{\prime }\right)}^2+2xy{y}^{\prime }–y^2=0\dots \left(1\right)$

For orthogonal trajectory, replace $y^{\prime }$ with $-\frac{1}{y^{\prime }}$ in (2), we get

${\left(\frac{y}{y^{\prime }}\right)}^2–2x\frac{y}{y^{\prime }}–y^2=0$

$⟹{\left(y{y}^{\prime }\right)}^2+2xy{y}^{\prime }–y^2=0$