Question

dy/dt= y-t^2 ..how would you get y(t)?

Answer 1

Are you wanting to integrate?

Answer 2

the problem asks me to use the graph of the solution y(t), and dy/dx is given

Answer 3

i think i have to integrate

Answer 4

Think you can upload the picture here some way?

Answer 5

http://math.sci.ccny.cuny.edu/document/show/2156

Answer 6

page 47, #18c

Answer 7

Yeah just integral both sides with respect to t

Answer 8

Yeah, let's first integrate to get back to the original
\[\int\limits_{}^{}y-t^2\]

So I guess \[\int\limits_{}^{}y - \int\limits_{}^{}t^2\]

Answer 9

i was wondering if i could do that...but for some reason i felt it was wrong

Answer 10

Then you will obtain equation with constant.

You were given \(y(0)=1\) so plug in y = 1 and t = 0 then solve for c

Answer 11

if you want to solve for y you will have to find the integrating factor

Answer 12

\[y'-y=-t^2 \\ y'+\color{red}{(-1)}y=-t^2 \\ \text{ the integrating factor is } e^{\int\limits_0^t -1 dx}=e^{-t}\]


\[e^{-t} y'+(-1)e^{-t}y=e^{-t}t^2 \\ (e^{-t}y)'=e^{-t}t^2\]

now you can integrate both sides

Answer 13

is that how i will find the original equation?

Answer 14

this is a first order, linear (non-separable) differential equation.
that is how it is classified, but to understand this, and understand it completely, all you need to know is:

\(\color{#000000}{ \small \displaystyle [1]\quad\quad \frac{d}{dx}e^{f(x)}=e^{f(x)}\times f'(x) }\) (via Chain Rule)

\(\color{#000000}{ \small \displaystyle [2]\quad\quad \frac{d}{dx}\left[f(x)g(x)\right]=f'(x)g(x)+f(x)g'(x) }\) (definition of Product Rule)

Answer 15

If you know these two things, then I can proceed to explaining how to solve for \(y(t)\) in such problems. So, do you, or not?

Answer 16

ok, yes i understand it