The particular solution of $e^{\dfrac{d y}{d x}}=2 x+1$ given that $y=1$ when $x=0$ is

Given the differential equation $e^{\dfrac{dy}{dx}} = 2x + 1$.

Taking the natural logarithm on both sides, we get $\dfrac{dy}{dx} = \ln(2x + 1)$.

Integrating both sides with respect to $x$, we have $y = \int \ln(2x + 1) dx$.

Let $u = 2x + 1$, then $du = 2 dx$, which implies $dx = \dfrac{1}{2} du$.

Substituting these into the integral, $y = \dfrac{1}{2} \int \ln(u) du$.

Using integration by parts, $\int \ln(u) du = u \ln(u) - u + C$.

Thus, $y = \dfrac{1}{2} (u \ln(u) - u) + C = \dfrac{1}{2} ((2x + 1) \ln(2x + 1) - (2x + 1)) + C$.

$y = \left(x + \dfrac{1}{2}\right) \ln(2x + 1) - x - \dfrac{1}{2} + C$.

Given $y = 1$ when $x = 0$, substituting these values: $1 = \left(0 + \dfrac{1}{2}\right) \ln(1) - 0 - \dfrac{1}{2} + C$.

$1 = 0 - 0 - \dfrac{1}{2} + C$, which gives $C = 1 + \dfrac{1}{2} = \dfrac{3}{2}$.

Substituting $C$ back into the equation: $y = \left(x + \dfrac{1}{2}\right) \ln(2x + 1) - x - \dfrac{1}{2} + \dfrac{3}{2} = \left(x + \dfrac{1}{2}\right) \ln(2x + 1) - x + 1$.

Answer: $y=\left(x+\dfrac{1}{2}\right) \log |2 x+1|-x+1$