Solution of differential equation xdy – ydx = Q represents

The correct option is: straight line passing through origin

Let us solve the given differential equation step-by-step to understand why it represents a straight line passing through the origin.

The given differential equation is:
$$xdy-ydx=0$$

We can rearrange this equation by separating the variables. Adding $ydx$ to both sides, we get:
$$xdy=ydx$$

Now, dividing both sides by $xy$ (assuming $x\ne 0$ and $y\ne 0$) to separate the variables $x$ and $y$:
$$\frac{dy}{y}=\frac{dx}{x}$$

Integrating both sides of the equation:
$$��\frac{dy}{y}=\int \frac{dx}{x}$$

The integration yields:
$$\ln \left|y\right|=\ln \left|x\right|+\ln \left|c\right|$$
where $\ln \left|c\right|$ is the constant of integration (with $c\ne 0$).

Using the logarithmic property $\ln \left(a\right)+\ln \left(b\right)=\ln \left(ab\right)$, we can write the right side as:
$$\ln \left|y\right|=\ln \left|cx\right|$$

Taking the exponential of both sides gives:
$$y=kx$$
where $k$ is an arbitrary real constant (incorporating the sign and value of $c$, and including $k=0$ which corresponds to the trivial solution $y=0$).

The equation $y=kx$ is the standard equation of a family of straight lines passing through the origin $(0,0)$, where $k$ represents the slope of the line.