Q.Verify that $y = kx$, where $k$ is an arbitrary constant, is a solution of the differential equation $x\dfrac{dy}{dx} = y$.

Differentiate the candidate solution and substitute both it and its derivative back into the differential equation.

The candidate solution is $y = kx$, where $k$ is an arbitrary constant.

Differentiating with respect to $x$: $\dfrac{dy}{dx} = k$ (since $k$ is a constant, its coefficient carries straight through).

Substitute into the left side of the differential equation $x\dfrac{dy}{dx}$: $x \cdot k = kx$.

The right side of the differential equation is simply $y$, which by the candidate relation equals $kx$.

Since $x\dfrac{dy}{dx} = kx$ and $y = kx$, both sides are identical — the equation $x\dfrac{dy}{dx}=y$ holds true for every value of $x$ and for every choice of the constant $k$, confirming $y=kx$ is indeed the general solution of this differential equation. …

A common error is substituting only the candidate function $y$ into the differential equation while forgetting to also compute and substitute its derivative $\dfrac{dy}{dx}$ — verification always requires BOTH the function and (at least) its first deri …