Question

solve the diff eq

Answer 1

\[x^2y'-4xy=x^7 sinx\]

Answer 2

@Zepdrix

Answer 3

how do we know which type of diff eq to use?

Answer 4

Mmm can you give me that link again?
That pdf. The one that talked about the matching "degrees".
Cause it looks like that's what we have going on here:

x^2 and xy are the same "degree".

Answer 5

Oh it's in my messages box :D nevermind.
one sec

Answer 6

http://www.math.hawaii.edu/~lee/calculus/DE.pdf

Answer 7

i was thinking we could use the first order diff eq where we have an integration factor

Answer 8

One sec im a little rusty XD
I think integrating factor is the right way to go.
Just need to do it on paper real quick hehe.

Answer 9

okie ;p

Answer 10

Ok ok ok I think I'm remembering :d
So ya... this is a first order linear differential equation.
Let's divide the x's off the first term and try an integrating factor,\[\large\rm y'-\frac4xy=x^5\sin x\]

Answer 11

So our D.E. is in this form:\[\large\rm y'+py=q\]
And what we want to do is multiply by some value u,\[\large\rm uy'+(pu)y=q\]such that pu is the derivative of u.
That way we have a perfect product rule in reverse,\[\large\rm uy'+u'y=uq\]Which we simplify using our product rule,\[\large\rm (uy)'=uq\]And do some integration.

So the derivative of u needs to be equivalent to pu.
That's what gives us all this magic.
\(\large\rm u'=pu\)

Doing some stuff will eventually get you to the realization that a substitution of the form:\[\large\rm u=e^{\int\limits p~dx}\]will suffice.

If that's too much detail, I apologize :P
We can just use that last part that I mentioned.

Answer 12

So ... recognize that the subtraction sign will be a part of our substitution,\[\large\rm y'+\color{orangered}{\left(-\frac4x\right)}y=x^5 \sin x\]

Answer 13

\[\huge\rm u=e^{\int\limits \color{orangered}{\left(-\frac4x\right)}~dx}\]

Answer 14

Can you figure out the u value for me? :D
Simplify it down.

Come on Macha! Lemme see you do it!

Answer 15

arghh wait its not posting

Answer 16

Hmm you're kinda close :)

Answer 17

oh yeh.. ;p

Answer 18

You can't apply your e^lnx=x thing until you've gotten the -4 out of the way.

Answer 19

Ok so that's our integrating factor.
Let's multiply through by x^-4

Answer 20

\[\large\rm \frac{1}{x^4}y'-\frac{4}{x^5}y=x \sin x\]Ok with that step?
Make sure you're getting the same thing.
(Don't forget to multiply the RIGHT SIDE by x^-4 as well.)

Answer 21

okie

Answer 22

If we have the right integrating factor,
then the left side should simplify to (uy)' by product rule in reverse,\[\large\rm \left(\frac{1}{x^4}y\right)'=x \sin x\]

Answer 23

You can check your work by applying product rule,\[\large\rm \left(\frac{1}{x^4}y\right)'\quad=\quad \left(\frac{1}{x^4}\right)'y+\left(\frac{1}{x^4}\right)y'\]Which gives us,\[\large\rm \left(\frac{1}{x^4}y\right)'\quad=\quad \left(-\frac{4}{x^5}\right)y+\left(\frac{1}{x^4}\right)y'\]So ya, good times.

Answer 24

So now integrate,\[\large\rm \int\limits\left(\frac{1}{x^4}y\right)'=\int\limits x \sin x\]

\[\large\rm \frac{1}{x^4}y=\int\limits x \sin x\]

Answer 25

Right side will require ummm .. integration by parts.
Hopefully you remember that from Calc 2,
because you'll be using it a lot in this class :D

Answer 26

You want to DESTROYYYYY the x, so that will be your u,
While the sinx will just cycle around, so that's our dv.

Then du = dx
and v = -cosx

And apply your IBP:
int(u dv) = uv - int(v du)

Answer 27

I'll write that information down again if it helps:
\[\large\rm u=x\]\[\large\rm dv = \sin x~dx\]and
\[\large\rm du=dx\]\[\large\rm v=-\cos x\]

Answer 28

\[\large\rm \int\limits x \sin x~dx\quad=\quad uv-\int\limits v~du\]\[\large\rm \int\limits x \sin x~dx\quad=\quad -x \cos x-\int\limits -\cos x~dx\]

Answer 29

\[\large\rm \int\limits\limits x \sin x~dx\quad=\quad -x \cos x+\sin x\]

Answer 30

So then,\[\large\rm \frac{1}{x^4}y=\int\limits\limits x \sin x\]gives us,\[\large\rm \frac{1}{x^4}y=-x~\cos x+\sin x + C\]And multiplying through by x^4 will give us our general solution,\[\large\rm y=-x^5~\cos x+x^4\sin x + x^4C\]

Answer 31

is this the final answer xD

Answer 32

Yes.

Answer 33

question how did this happen again?
was it bc the integral canceled with the y'?

Answer 34

The prime (derivative) was being applied to the whole thing,
not just the y. So careful with that :O it's not just a y'.

But yes, integration is 'anti-differentiation'.
So you're taking the anti-derivative of a derivative :)
They are inverses of one another and essentially "cancel out"

Answer 35

oh okie

Answer 36

@Zepdrix lmaoo

Answer 37

I've never seen the movie, so the reference is a little lost on me lol
I've always enjoyed Big Vikings comedic style since they took over YoWorld tho x'D

Answer 38

yeh lmaooo the tie though hahaha