\(y= 1+\int_0^x y(t) dt\) (1)

First we find the first-order differential equation by differentiating both sides with respect to x

\(dy/dx= d/dx(1+\int_{0}^{x} y(t)dt)\) (2)

As according to fundamental theorem of calculus

\(d/dx \int_{a}^{x} f(t)dt= f(x)\)

So we can write (2) as

\(dy/dx= y(x)\ or\ dy/dx=y\), \(y'=y\)

Now we find the initial conditions for y

The equation is of the form

\(y= f(c)+\int_{c}^{x} g(t)dt\)

Now on comparing this with equation (1) we get

\(f(0)=1\ and\ c=0\)

So, the initial condition for y is \(y(0)=1\)

Therefore the first order differential equation is \(y'=y\) and initial condition for y is \(y(0)=1\)