Let F (x, y)be the curve passing through the origin.
At point (x, y), theslope of the curve will be
According to the given information:

This is a linear differential equation of the form:

The general solution of the given differential equation isgiven by the relation,

Substituting in equation (1), we get:

The curve passes through the origin.
Therefore, equation (2) becomes:
1 = C
⇒ C = 1
Substituting C = 1 in equation (2), we get:

Hence, the required equation of curvepassing through the origin is