Solution of dy/dx – y = 1 y(0) = 1 is given by

The given first-order linear differential equation is:
$$\frac{dy}{dx}-y=1$$
with the initial condition:
$$y\left(0\right)=1$$

Step 1: Separate the variables
We can rewrite the differential equation by moving the term $y$ to the right-hand side:
$$\frac{dy}{dx}=y+1$$
Now, divide both sides by $(y+1)$ and multiply by $dx$ to separate the variables $y$ and $x$:
$$\frac{1}{y+1}dy=dx$$

Step 2: Integrate both sides
Integrate both sides of the equation with respect to their respective variables:
$$\int \frac{1}{y+1}dy=\int dx$$
Integrating, we get:
$$\ln |y+1|=x+C$$
where $C$ is the constant of integration.

Step 3: Solve for y
Exponentiate both sides to eliminate the natural logarithm:
$$|y+1|=e^{x+C}$$
Using exponent rules, this can be written as:
$$y+1=A{e}^x$$
where $A=\pm e^C$ is a new constant.
So, the general solution is:
$$y=A{e}^x-1$$

Step 4: Apply the initial condition to find A
We are given that $y\left(0\right)=1$. Substitute $x=0$ and $y=1$ into our solution:
$$1=A{e}^0-1$$
Since $e^0=1$:
$$1=A-1$$
Solving for $A$:
$$A=2$$

Conclusion
Substituting $A=2$ back into the equation yields the particular solution:
$$y=2e^x-1$$
Which corresponds to the option:
$$y=2e^x-1$$