Question

Solve the IVP.
y'=3t^(2)-y y(0)=1 [0,1]

Answer 1

The is a linear diff equation
y'+y=3t^2

multiply by v on both sides
vy'+vy=v3t^2
where v is the integrating factor

Answer 2

you want to choose v such that you have the left hand side is (vy)'=vy'+v'y.
So what does v' need to be in order for vy'+vy to be the same?

Answer 3

I dont understand what you're trying to say

Answer 4

freckles is saying it is a linear differential equation

Answer 5

and he/she is trying to get you to remember the product rule

Answer 6

where v and y are functions of x
(vy)'=vy'+v'y
you have on the left hand side of vy'+vy=3t^2
vy'+vy

In order for vy'+v'y=vy'+vy
v' needs to be the same as v
This is how we will choose v
v=v'
assuming y and v are functions of x
we have
v=dv/dt

this equation can be solved by separation of variables

Answer 7

dt=dv/v
now integrate

Answer 8

however if you aren't used to deriving your own integrating factor
and you have become reliant upon the form \[v=e^{ \int\limits_{}^{}p(x) dx } \text{ where you have written the differential equation } \\ y'+p(x)y=q(x)\]

Answer 9

this is not the answer this just the integrating factor
this v is what you need to multiply both sides so that
you have
(vy)'=vq
Then you integrate both sides giving you
\[vy=\int vq dx +C \]

Answer 10

what do i do with the intervals?

Answer 11

with this interval [0,1]

Answer 12

we will get there after we find y

Answer 13

also y(0)=1 is an ordered pair

Answer 14

we will use this to find C

Answer 15

also you had functions of t not x
whatever we will say the variable x represents the variable t

Answer 16

am i doing this good so far y+dy=3t^2 dx

Answer 17

Ok I don't think you understand...
Let's see if you can following this example:
\[\frac{1}{t}y'+y=t^2 \]
y is a function of t
so first write as
\[y'+ty=t^3\]
I just multiply t on both sides
dt/dt=1
so we already have (ty)'=t^3
integrate both sides with respect to t
we have
ty=t^4/4+C

Answer 18

Here is a harder question
\[\frac{1}{t}y'-y=t^2 \]
write in the y'+py=q form
\[y'-ty =t^3\]
multiply both sides by v so that we can sue product rule on left side
we get to choose v to be anything (well except 0)
\[vy'-vty=vt^3 \]
we want to choose this v so we can write the left hand side as vy'+v'y
so this means we want -vt=v'
so this means we first need to find this v from this equation before we continue
we need to know what we multiplied by
\[-vt =\frac{dv}{dt} => -t dt = \frac{dv}{v} => -\frac{t^2}{2}=\ln|v| \]
So \[v=e^{\frac{-t^2}{2}}\]
so we have
\[(vy)'=e^{\frac{-t^2}{2}} \cdot t^3\]

The integrate both sides you can go ahead and replace the other v as well

Answer 19

You can skip all of this though and use the formula
but formulas are no fun

Answer 20

all of the first order linear differential equations
that is y'+py=q can be solved in the same way

Answer 21

first find \[v=e^{ \int\limits_{}^{} p(x) dx }\]
This is called the integrating factor once you put your equation in this form y'+py=q

Answer 22

Then integrate both sides of (vy)'=vq

Answer 23

which will lead to \[vy= \int vq dx +C\]

Answer 24

but I only recommending using this formula if you actually understand how it works
but even then I never like to memorize this formula
because it is fun getting there without it

Answer 25

yea im so confused don't i need to use Euler's method?

Answer 26

I find this way easiest mostly because I don't remember the way.
I would have to go back and do review.
I think @rational is watching this like a hawk though. So if he doesn't chime in, I will after I do a quick review.

Answer 27

I don't remember the other way*

Answer 28

There is also another way where you can guess what y is based on that 3t^2 thing

Answer 29

but i need a few minutes or maybe 30 for euler's method

Answer 30

so we have f(0,1) <--this represents slope at (0,1)

Answer 31

and y'=3t^2-y

Answer 32

so y'=3(0)^2-1
so the slope at (0,1) is -1
so f(0,1)=-1

Answer 33

The next step is multiply by a step size h
hf(y_0)=hf(1)=h(-1) (we will let h=1)
so hf(y_0)=-1
where y_0=-1 and t_0=0

does this look familiar so far?

Answer 34

then keep evaluating :
y_n=y_(n-1)+hf(y_(n-1))

Answer 35

but usually i think you use this method to approximation other values

Answer 36

Could be wrong about that.

Answer 37

hey @mony01 why do you want to use a numeral method when you can solve it analytically ?

Answer 38

oops that y_0 is suppose to be 1 (that was a type-0)

Answer 39

euler's numerical method gives you a set of points using which you can fit a polynomial/exponential curve.. but i feel you should think of numerical methods only when its not possible to solve it other ways

Answer 40

Also, that integrating factor idea was first occurred to Euler only.
So please be clear on what you mean by euler's method because Euler is there everywhere. "euler's method" is not a standard phrase. Do you mean Euler's numerical method ?

Answer 41

okay this is the original question for the problem, now im confused on what formula to use. @rational @myininaya
Solve the IVP numerically on the suggested interval, if given, using various step sizes. Compare with values of exact solutions when possible.

Answer 42

http://www.swarthmore.edu/NatSci/echeeve1/Ref/NumericInt/Euler1.html
This is why I think euler's method is used to approximate

Are we trying to appoximate a certain value
like y(1/2) or something?

Answer 43

I dont really know well the chapter from this problem is all about Euler