Integral Equations $y(x) = 2 + \int_x^2 \left[ t - ty(t) \right] dt$

Question

My equations is:

$$ y(x) = 2 + \int_x^2 \left[ t - ty(t) \right] dt $$

What I am not understanding is if I take the derivative of both sides I don't understand how my book gets x-xy on the right hand side. To my understanding if I take the derivative of an integral, it is almost like it is "cancelling" it out. So why wouldn't it be $x-xy(x)$ or $x-yx^2$?

Answer 1

HINT

Let $F(t)$ be the anti-derivative of $f(t)$, then $$ \frac{d}{dx} \int_x^2 f(t) dt = \frac{d}{dx} \left[ F(2)-F(x)\right] = -f(x). $$

Answer 2

The fundamental theorem of calculus says (skipping the assumptions here) that if $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$.

In the above, the lower limit of the integration is the variable, so $F(x) = \int_x^a f(t) dt = - \int_a^x f(t)dt$, so we have $F'(x) = - f(x)$.

Explicitly, $F(x) = \int_x^2 (t-t y(t)) dt$, so $F'(x) = -(x-x y(x))$.

In the question, the right hand side is $2+F(x)$, so the derivative of the right hand side is $-(x-x y(x))$.