Question

Differential Equations home work:

Answer 1

So, there is no way for you to solve this.

Answer 2

Find solution to the IVP and give in standard conic form \[\left( y ^{2}+16 \right)dx + \left( 3y-xy \right)dy = 0 \]

Answer 3

Actually h/o

Answer 4

y(0) = 2sqrt(5)

Answer 5

\[y(0)=2 \sqrt5\]

Answer 6

Yeah i was trying to do that but fluttered it up haha

Answer 7

Ready?

Answer 8

Yeah drop some multivariable knowledge on me

Answer 9

\[(y^2+16)dx+y(3-x)dy=0 \implies -y(3-x)dy=(y^2+16)dx \implies \frac{-y dy}{y^2+16}=\frac{dx}{3-x}\]
\[-\int\limits \frac{y dy}{y^2+16}=\int\limits \frac{dx}{3-x} \implies -\frac{1}{2}\ln(y^2+16)=\ln(3-x)+C\]
Find C. Then solve for Y.

Answer 10

There isn't any multivar here :P

Answer 11

Well I certainly was looking at it the wrong way lol. Just out of curiosity can you integrate (y^2+16)dx at all like it is? You can't can you?

Answer 12

You could. \[\int\limits y^2+16 dx=xy^2+16x+C\]

Answer 13

Oh, well that's exactly what I wrote down but I figured that wasn't the right way to solve this problem

Answer 14

Because:
\[\frac{\partial}{\partial x}(xy^2+16x+C)=y^2+16\]

Answer 15

Its not. The way I did is the right way^^ You can't just run a derivative across a sum of differentials like that. Integration is a linear operator meaning:
\[\int\limits(a+b)=\int\limits a+\int\limits b \]
That does not imply:
\[a+b=0 \implies \int\limits a+\int\limits b=\int\limits 0\]
You can't just do that.

Answer 16

got ya

Answer 17

Thats like going:
\[a+b=2 \implies \sin(a)+\sin(b)=\sin(2)\]
Which is not true in general.

Answer 18

So my next question is how am i supposed to solve this with all those logs?

Answer 19

It's like Ln(3) + (1/2)(ln(y^2+16)) = C

Answer 20

Let y=2sqrt(5) since you plugged in zero already for x.
Then just use ln properties

Answer 21

That works out SO fluttering nice bro.

Answer 22

uhm ... idk how exactly ... i'm still getting something like ln(3)= -ln(36) / 2 + C

Answer 23

Well:
\[\ln(3)+\frac{1}{2}\ln(36) \implies \ln(3)+\ln(36^{\frac{1}{2}}) \implies \ln(3)+\ln(6) \implies \ln(18)\]

Answer 24

Now i have to write that in "standard conic form" wtf kind of conic involves log functions

Answer 25

Thats not what they mean by conic form.

Answer 26

Lol... Yeah I realized that after I typed it. Do you remember what conic form is though i dont remember what it's supposed to look like

Answer 27

Its the same as standard whatever: http://www.uoguelph.ca/~math2080/pdf/Lectures/w6.pdf

Answer 28

How can the standard form and standard conic form mean the same when in standard form it is like y' = b(t) whatever

Answer 29

Theres no y' in the form where it has C in it lol

Answer 30

Hey just to check my algebra could you put that in explicit form and show me?

Answer 31

−12ln(y2+16)=ln(3−x)+C

Answer 32

-(1/2)ln(y^2+16) = ln(3-x) + C that is

Answer 33

You want to solve that for y?

Answer 34

Yeah

Answer 35

I'm doin something wrong lol. I can't graph it properly. I either get something outrageously small or I get a line

Answer 36

Lol this is a tad frustrating I have to admit. Theres nothing more that I hate than doing Honeycutt home work... I'd EASILY take Mrs. Wallins home work