Question

# Write an equivalent first-order differential equationand initial condition for yy= 1+int_0^x y(t) dt

First order differential equations

Write an equivalent first-order differential equationand initial condition for y $$y= 1+\int_0^x y(t) dt$$

2020-11-24

$$y= 1+\int_0^x y(t) dt$$ (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 $$y(0)=1$$