Question

DIFFERENTIAL EQUATIONS:
Question:
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36/MATA36-A4Attempt2Q.png

Attempt:
(1)
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36/MATA36-A4-Attempt2a.png
(2)
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36/MATA36-A4-Attempt2b.png

What did I do wrong here?

Answer 1

@eliassaab
@saifoo.khan
@AccessDenied
@Zarkon
@dpaInc
@TuringTest
@amistre64
@Mertsj
@Hero
@zepp
@lgbasallote

Anyone?

Answer 2

Oh so you can use either logs or ln's?

Answer 3

Anyways, I will compare

Answer 4

Where did the:
\[\int\limits_{}^{}(8+y^2)^{1/2}dy\]

go?

Answer 5

Is your answer fully reduced? 117 /12 is reducible (117 and 12 are divisible by 3)

Answer 6

Oh you are saying that I may have got it wrong because I did not reduce fully?

Answer 7

It's possible. I think a lot of programs would like the simplest form version. D:
I got the same answer, following steps in some notes.

Answer 8

But with a negative sign?

Answer 9

I got it with a negative sign

Answer 10

@AccessDenied did you get -117/12, or 117/12

Answer 11

I get negative when i keep the constant on the x-side (as seems to be conventional).
I reduced it to -39/4 though

Answer 12

Still wrong... Darn it.

Answer 13

Wait a sec, I think I did put it on the y-side. >.>
It "should" go on the x-side / left side, which would make the c-value I got positive. I'm just double checking though.

Answer 14

No way... Really? So a convention really does make a difference

Answer 15

This was my work, in the attachment.
I'm not 100% certain, but the work at least seems correct to me. o.o

Answer 16

Do you use tablet? That looks pretty legible

Answer 17

Laptop's touchpad. It's sometimes hard to control (like it'll start getting fidgety), but it seems to come out nicely.

Answer 18

39/4, positive is still wrong.

Answer 19

I just checked.

Answer 20

Hmm, I'm just not sure then. :(

Answer 21

Then its alright man. You have done what you could.

Answer 22

Sorry I copied the problem wrong. I put log[1+y^2] instead of sqrt(1+y^2).
The method is still the same.

Answer 23

I will fix it and post it.

Answer 24

\[
\int x \log (x) \, dx=\frac{1}{2} x^2 \log
(x)-\frac{x^2}{4}\\
\int y \, dy=\frac{y^2}{2}\\

\int y \sqrt{y^2+8} \, dy=\frac{1}{3}
\left(y^2+8\right)^{3/2}\\

\int \left(\sqrt{y^2+8} y+y\right) \,
dy=\frac{1}{3}
\left(y^2+8\right)^{3/2}+\frac{y^2}{2}\\

\frac{1}{2} x^2 \log
(x)-\frac{x^2}{4}=\text{C1}+\frac{1}{3}
\left(y^2+8\right)^{3/2}+\frac{y^2}{2}\\

\text{C1}=-\frac{39}{4}\\

\frac{1}{2} x^2 \log
(x)-\frac{x^2}{4}=-\frac{39}{4}+\frac{1}{3}
\left(y^2+8\right)^{3/2}+\frac{y^2}{2}

\]

Answer 25

Oh, wait.. the actual 'solution' is where you plug in the C1? Not just the C1 itself...
Well, that would make sense.

Answer 26

Ok that makes sense because the solution of a differential equation is a function right...

Answer 27

The convention thing I mentioned only changes the actual C-value, since if you have the C on one side, it is one way, but equivalently if you subtracted it from both sides to the other side, it becomes a negative C.
In the actual solution though, it always checks out.

Answer 28

I found out what I messed up on. I am suppose to express this as y = (ie. write out it as a function, not a value)

Answer 29

But in this case, I cannot do that so I will write out both the LHS = RHS as my solution

Answer 30

With the value of C.

Answer 31

OMG! It is correct! You have to write out both sides of the function!

Answer 32

Nice. :D
I'll have to keep this in mind for when I start DE. lol

Answer 33

This is how the solution is expressed IMPLICITLY. @AccessDenied.

Answer 34

So basically, it would be:
ln(x)*x^2/2 - x^2/4 = -39/4 +y^2/2 + (1/3)*(8 + y^2)^{3/2}

That is the answer defined implicitly. Here, it is impossible to define explicitly in terms of y or x.

Answer 35

I think

Answer 36

Yeah, I'm not seeing any way either. the square root has that extra addition tacked on and the x has that logarithm.

Answer 37

Do you know what happened here? Look at this below:
\[|y| = e^{\frac{x^3}{3} + C} \]

Then, they got:
\[y = \pm e^{\frac{x^3}{3} + C}\]

How is that possible?

Answer 38

They went from absolute value to + or -. How?

Answer 39

Because the absolute value makes the expression always-positive, so we have to consider the case where the original y is negative to remove them.

Answer 40

Like, if y = -2 OR 2, |y| = 2

Answer 41

Got it. Thank you.

Answer 42

You're welcome. :)