naivlingr

2020-11-23

Write an equivalent first-order differential equationand initial condition for y $y=1+{\int }_{0}^{x}y\left(t\right)dt$

liannemdh

Expert

$y=1+{\int }_{0}^{x}y\left(t\right)dt$ (1)
First we find the first-order differential equation by differentiating both sides with respect to x
$dy/dx=d/dx\left(1+{\int }_{0}^{x}y\left(t\right)dt\right)$ (2)
As according to fundamental theorem of calculus
$d/dx{\int }_{a}^{x}f\left(t\right)dt=f\left(x\right)$
So we can write (2) as
, ${y}^{\prime }=y$
Now we find the initial conditions for y
The equation is of the form
$y=f\left(c\right)+{\int }_{c}^{x}g\left(t\right)dt$
Now on comparing this with equation (1) we get

So, the initial condition for y is $y\left(0\right)=1$
Therefore the first order differential equation is ${y}^{\prime }=y$ and initial condition for y is $y\left(0\right)=1$

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