I am looking for a curve such that curvature at point should be proportional to $x$ coordinate.
Let us say that the curve is given by (x,f(x)). Then the curvature is
\begin{eqnarray} \kappa = \frac{y''(x)}{(1+y'(x)^2)^{\frac{3}{2}}} \end{eqnarray}
And if curvature is to depend on $x$ linearly, then
$$\kappa = \mu + \lambda x$$, where $\lambda, \mu$ are constants. I know that when $\lambda$ is equal to zero the solution is circle. But what is the general solution of the problem?
Assuming $y'(x) \ll 1$, the solution will be a cubic polynomial. This does not sound a nice solution.
Using the notation for intrinsic curves, $s$ for arc-length and $\psi$ for tangential angle, where $\tan\psi=\frac{dy}{dx}$, curvature is given by $\frac{d\psi}{ds}$.
So we have $$\frac{d\psi}{ds}=kx=\frac{d\psi}{dx}\frac{dx}{ds}=\frac{d\psi}{dx}\cos\psi$$
$$\implies\frac 12kx^2+c=\sin\psi=\frac{\tan\psi}{\sec\psi}=\frac{\frac{dy}{dx}}{\sqrt{1+\left(\frac{dy}{dx}\right)^2}}$$
Solving this DE would give the general solution.
So, for example, in the case where $c=0$ and $k=2$, this boils down to finding $$\int\frac{x^2}{\sqrt{1-x^4}}dx$$ which is expressible in terms of elliptic functions.