Question

Going over some basic Differential Equations. Posted below momentarily.

Answer 1

Solve the differential equation.\[\frac{y}{\sqrt{4-x^2}}=y^2+5\]

Answer 2

@dan815 , could you help me out with this? I really don't remember how to do this...

Answer 3

Is that \(y'\) on the left?

Answer 4

Sorry, I forgot an entire term; the lefthand side is multiplied by y prime.

Answer 5

No problem. The equation is separable:
\[\frac{y}{\sqrt{4-x^2}}y'=y^2+5\]
Separating yields
\[\frac{y}{y^2+5}dy=\sqrt{4-x^2}~dx\]

Answer 6

The left can be integrated with a substitution, the right with a trigo sub.

Answer 7

Alright, gonna take a shot at the rest from here, one sec. Thank you so far.

Answer 8

yw

Answer 9

\[u = y^2+5, \ du = 2y \ dy; \ \ \ \]

Answer 10

Ugh, trig subs....is this the thing where.......one sec, lol. Gotta think about trig identities for a second again.

Answer 11

Could x be substituted for either 2sin(theta) or 2cos(theta), if I recall correctly?

Answer 12

In this case, yes, either works. The sine sub is usually taken to avoid working with the pesky negative the cosine gives.

Answer 13

Hey, I'm not too familiar with LaTeX, do you know how I "turn off" the \rm statement in the middle of a sentence? e.g. I want to type upright in text using it, but then I want to put a math formula in the sentence. I'll do it either way, but I want to figure that out.

Answer 14

\text{ }

Answer 15

i don't know
maybe i misunderstood

Answer 16

\[\frac{y}{y^2 +5}\frac{dy}{dx}=\sqrt{4-x^2}; \ \ \ \frac{1}{2}\int\limits_{}^{}\frac{du}{u}du=\int\limits_{}^{}\sqrt{4-4 \sin(\theta)}dx\]

Answer 17

I think OP is wondering how to type something like \(\sum\) in-line with text? In which case you'd wrap your code in backslash-parentheses, as in `\( <code> \)`.

Answer 18

Let me try it out. \[\rm Let \( swag \)\]

(Lol, didn't work)

Answer 19

something like this:
\[\frac{ \text{ see a fish }}{ \text{ be a fish } } \text{ is a fraction }\]

Answer 20

But yeah, I want math formulas with slanted, italicized variables and all, in the middle of a sentence in LaTeX.

Answer 21

\[\frac{1}{2}\ln(y^2+5)+c_{1}=\int\limits_{}^{}2\cos(\theta)dx\]
I forget from here how you change the right-hand's variable of integration to d-theta in a situation where you use trig substitution.

Answer 22

AHH yeah, I got it, it was because I was using \rm, I know you mentioned \text{}, but I've never used it, thank you~!

Answer 23

Maybe like this?\[\text{Let }swag\]
A few ways you can do it.
`\[\text{Let }swag\]` (as above)
`\[\text{Let }\emph{swag}\]` (as below)
\[\text{Let }\textit{swag}\]
The differences are slight, mainly with the kerning and scaling of the letters.

Answer 24

Sorry, `\emph` doesn't work on this site... I actually had used `\textit{...}`.

Answer 25

(Lol)

Alright, but yeah, moving forward with the problem, how do I deal with that differential being dx? I forgot how this works.

Answer 26

http://openstudy.com/study#/groups/LaTeX%20Practicing!%20%3A)
this is a good group to learn latex in
people have made a lot of threads on latex tutorials

Answer 27

You set \(x=2\sin\theta\), which gives what differential for \(x\)? \(dx=\cdots\)

Answer 28

Oh, derp. Yeah, I gotchu.

Answer 29

>The fact that I didn't realize that

feelsbadman.jpg

\[\frac{1}{2}\ln(y^2+5)+c_{1}=\int\limits_{}^{}2\cos(\theta)[2\cos(\theta)d \theta]\]

Answer 30

I'm moving slowly on this because the whole form of this thing looks like it's about to play a trick on me, the integrand being identical to the differential element, lol....alright. Integrating the RHS:

Answer 31

\[4\int\limits_{}^{}\cos^{2}(\theta)d \theta = 2\int\limits_{}^{}(1+\cos(2 \theta)d \theta\]

Answer 32

\[2 \int\limits_{}^{}1+\cos(2 \theta))d \theta = 2\bigg[\theta+\frac{\sin(2 \theta)}{2}\bigg]\]

Answer 33

\[2 \theta + \sin(2 \theta)+c_{2}=\frac{1}{2}\ln(y^2+5)+c_{1}\]

Answer 34

(Sloppy introduction of c_{2}, whoops.)

Answer 35

Uhh, something subbing x stuff where theta is.

Answer 36

I like to draw a right triangle for my trib substitution
and use it as a reference later to put back in terms of what I started with

Answer 37

also sin(2u)=2sin(u)cos(u)

Answer 38

Yeah, I'm just being a dummy and forgetting how to do it algebraically as well.

Answer 39

One sec.

Answer 40

\[2\sin(\theta) = x; \ \ \ \sin(\theta) = \frac{x}{2}.\]
Can set up that reference triangle from there, one moment.

Answer 41

you only really need that triangle for the cos(theta) part

Answer 42

:|

I didn't need it for the cos(theta) part ironically (personally)-but I can see it being useful if I got a spaghetti mess of different trig functions from the original. Alright, either way, need to use the identity you mentioned.

Answer 43

\[2\theta + \sin(2 \theta) = 2\sin (\theta) \cos (\theta)\]

Answer 44

\[2 \theta+\sin(2 \theta)=2 \theta+2 \sin(\theta)\cos(\theta)\]

Answer 45

Theta by itself is:\[\sin^{-1}(x/2)\]

Answer 46

This is looking stupidly more complex than I expected for just a run-of-the-mill DE, I sure hope I didn't make a mistake somewhere.

Answer 47

\[\sin^{-1}(x/2)+x\frac{\sqrt{4-x^2}}{2}\]

Answer 48

\[\sin^{-1}(x/2)+\frac{x \sqrt{4-x^2}}{2}=\frac{1}{2}\ln(y^{2}+5)+C\]

(Just gonna collect together those arbitrary constants into one, I think I'm allowed to do that, but not 100% sure.)

Answer 49

Now what do I do?

Answer 50

that looks good

Answer 51

Well, don't I have to solve for something, or....something? Lol. I don't remember how to solve DE's like this, but I feel like I have to express the solution in some format other than this.

Answer 52

@wio , do I stop here, or does this look good?

Answer 53

I think that form is find to express the answer

but you can try to solve for y
just so you know it isn't always easy or possible to solve for y

Answer 54

fine*

Answer 55

Yeah, that's why, I feel quite a bit like (this was a sample final exam question) we would have to very explicitly express y as a function of x for the solution.