A particle is moving with a velocity v = K(yi^+xj^), where K is a constant. The general equation for its path is:

Question

A particle is moving with a velocity $\vec{v}=K(y\hat{i}+x\hat{j})$, where K is a constant. The general equation for its path is: 

(1) xy = constant 

(2) y² = x² + constant 

(3) y = x² + constant 

(4) y² = x + constant

Answer 1

Ans: (2) y² = x² + constant

Sol: ${dx\over dt}=ky, {dy\over dt}=kx$

Now, ${dy\over dx}={{dy\over dt}\over{dx\over dt}}={x\over y}$

$\implies ydy=xdx$

Integrating both side 

$y^2=x^2+c$