Question

Write an equivalent first-order differential equation and initial condition for y.

Answer 1

\[y=-1\int\limits_{0}^{x}(2t-y(t))dt\]

Answer 2

Do you have any ideas so far?

Answer 3

I'm as far as y'=

Answer 4

I just need some direction on how to go about working this problem.

Answer 5

What is first-order? I think I can do the integral.

Answer 6

First-order means the equation only contains the function and its first derivative.

Answer 7

So you are correct to take the derivative of both sides, the right-hand side should become a function of y and t and no extra higher order derivatives.

Answer 8

derivative, or integral?

Answer 9

Here:
\( \displaystyle y = - \int_{0}^{x} \left(2t - y(t)\right) \ dt \)
If we integrated both sides, we would not have the first derivative anywhere. Instead, derivative of both sides:
\( \displaystyle y' = \dfrac{d}{dx} \int_{0}^{x} \left(2t - y (t) \right) \ dt \)
Note the derivative with respect to x because the right-hand integral is a function of x when it is evaluated.

Answer 10

ahh, I see! Thanks!!

Answer 11

derive both sides, and my function is on the right.

Answer 12

Yup. The right-hand side will end up leaving the function y(x).
The d/dx of integral might be interesting to evaluate if you do not recall the first fundamental theorem of calculus, but if you treat integral of y(t) dt = Y(t) such that d/dt Y(t) = y(t), you can still evaluate by doing the integral first.

Answer 13

y'(x) is f(x,y(x))

Answer 14

so y'=2x-y

Answer 15

and the initial condition is y(0)=-1

Answer 16

Thanks for the help. I just needed some help jogging my memory.

Answer 17

that is about correct, I just saw I missed the negative sign in front when I wrote the derivative of both sides:
\( \displaystyle y' = \color{red}{-} \frac{d}{dx} \int_{0}^{x} 2t - y(t) \ dt \)
which was my bad. your right-hand side is just -1(2x - y) = y - 2x

Answer 18

although i don't see where you found y(0) = -1?
although the -1 is multiplied on, we have an integral from 0 to 0 (covers no area).

Answer 19

just to emphasize the substitution:
\( \displaystyle y \color{green}{(x)} = - \int_{0}^{\color{green}{x}} \left( 2t - y(t) \right) \ dt \)
x=0:
\( \displaystyle y(\color{red}{0}) = - \int_{0}^{\color{red}{0}} \left( 2t - y(t) \right) \ dt \)

Answer 20

I'll have to get back to this later. I have a 3 year old climbing me! Thanks again.

Answer 21

glad to help! and good luck. :p

in the end i do end up with y' = y - 2x and y(0) = 0. (I'll be getting off soon so I'll leave it here for any additional assistance).