Write an equivalent first-order differential equationand initial condition for y y=1+∫0xy(t)dt

$y=1+{\int }_0^{x}y(t)dt$ (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
$dy/dx=y(x)ordy/dx=y$, $y^{\prime }=y$
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
$f(0)=1andc=0$
So, the initial condition for y is $y(0)=1$
Therefore the first order differential equation is $y^{\prime }=y$ and initial condition for y is $y(0)=1$