Question

Differential Equations home work

Answer 1

Okay, now we can talk.

Answer 2

\[\frac{dy}{dx} = \frac{x ^{2}+5y ^{2}}{2xy}\]

with y(1) = \[\sqrt{?}\]

Answer 3

y(1) = sqrt(3)

Answer 4

What change of variable did you use?

Answer 5

I ended up with v+xv' = \[\frac{x ^{2}+5(xv)^{2}}{2x^{2}v}\]

Answer 6

hmm you can try by taking \(y =v*x\)

Answer 7

after that i had v+xv' = (1+5v^2)/(2v) ? is that right ?

Answer 8

\[\frac{dy}{dx} = \frac{dv}{dx}*x + v\]
\[ \frac{dv}{dx} + v = \frac{1 + 5v^2 }{2v}\]
\[\frac{dv}{dx} = \frac{1+5v^2}{2v}-v\]
\[\frac{2v}{1 + 5v^2 -2v^2} dv= dx\]

Answer 9

Now Integrate them :D

Answer 10

Where did your x go? you have just dv/dx? I thought it should be x(dv/dx)

Answer 11

\[\int \frac{2v}{1 + 3v^2}dv=\int d x\]
\[t = 1 + 3v^2 \]
\[\frac{dt}{6v} = dv\]
\[\frac{1}{3}*\int \frac{1}{t} dt = \int dx \]

Answer 12

hmm .... \[\frac{x ^{2}+5(xv)^{2}}{2x^{2}v}\]\[\frac{\cancel{x}^2(1 + 5v^2)}{\cancel{x^2}(2v})\]

Answer 13

is that what you were talking about ?

Answer 14

I understand that part but i thought y' = xv'(x)

Answer 15

y' = v+xv' **

Answer 16

\[y = v*x\]
\[\frac{dy}{dx} = (\frac{d}{dx}v)* (x) + (\frac{dx}{dx})* v \]

Answer 17

\[\frac{1}{3}*\log t + C_1 = x + C_2\]

Answer 18

\[2xy y'-5y^2=x^2\]

we need

(y^2)'=2yy'

so look we have
\[x (2yy')-5y^2=x^2\]

so we need
multiply by v(x)>0

\[vx(2yy')-5vy^2=vx^2\]

we need (vx)'=-5v
v'x+v=-5v
v'x=-6v
\[\frac{dv}{dx} x=-6v => \frac{1}{v} dv=-\frac{6}{x} dx\]
integrate both sides
\[\ln(v)=-6\ln|x|+k_0\]
\[\ln(v)=-6 \ln|x|+k_0, let k_0=0 =>\ln(v)=-6 \ln|x| =>v=x^{-6}=\frac{1}{x^6}, x \neq 0 \]
so we have
\[\frac{1}{x^6}x(2yy')-5\frac{1}{x^6}y^2=\frac{1}{x^6}x^2 \]
\[\frac{1}{x^5}(2yy')-\frac{5}{x^6}y^2=\frac{1}{x^4} \]
\[(\frac{1}{x^5}y^2)'=\frac{1}{x^4}\]
integrate both sides

Answer 19

\[t = 1 + 3v^2\]\[y = xv\]\[v=\frac{y}{x}\]
\[t = 1 + 3*\frac{y^2}{x^2}\]

Answer 20

\[\frac{1}{x^5}y^2=\frac{x^{-4+1}}{-4+1}+K \]

Answer 21

\[\frac{y^2}{x^5}=\frac{x^{-3}}{-3}+K\]
\[\frac{y^2}{x^5}=\frac{-1}{3x^3}+K\]

Answer 22

You are So Good with \(\LaTeX\) !!!

Answer 23

\[y^2=\frac{-x^5}{3x^3}+Kx^5 =\frac{-x^2}{3}+Kx^5 \]
\[y=\pm \sqrt{\frac{-x^2+3Kx^5}{3}}=\pm \sqrt{x^2}\frac{\sqrt{3Kx^3-1}}{\sqrt{3}}=\pm x \frac{\sqrt{3Kx^3-1}}{\sqrt{3}} \]

Answer 24

myininaya <3 lol

Answer 25

teach me your ways! D:

Answer 26

lol

Answer 27

i tried to force the linear method because its all i can remember lol

Answer 28

myininaya I was just saying to malevolence that i dont understand at all wtf she's doing lmao... I haven't been taught that yet =/

Answer 29

Thanks for the help though guys I'm starting to understand it

Answer 30

how to solve first order linear differential equations?

Answer 31

you haven't been taught that?

Answer 32

Nah not yet, the fall semester has just started and she hasn't actually taught us how to solve the equations yet, just how to do initial value problems and a few other things

Answer 33

omg she should really teach you how to solve something in this form first:
\[y'+p(x)y=q(x)\]

Answer 34

its really the most basic differential equation you can solve by sides any that you can solve for separation of variables

Answer 35

Yeah.... honestly i'm not sure about this teacher. Idk how i'm gonna end up doing because she gives really hard work and her lecture is minimal in the help department

Answer 36

It's a 2 hour long class... she's from upstate new york and talks incredibly fast... the class starts and she starts talking and doesn't stop till her 2 hours are up

Answer 37

lol

Answer 38

ask her to give you guys a review how to solve first order linear differential equations

Answer 39

are you guys scared of her?

Answer 40

Seriously i wish i could ask questions but she DOESN'T give time to ask questions in class

Answer 41

office hours?

Answer 42

she has her tablet computer with her stuff on the projector... she starts teaching and looks down at her tablet and basically doesn't lift her head for 5 minutes at a time. It's awful =/

Answer 43

Yeah that's about the only time I could possibly get to her to ask her questions but her office hours are the times that i'm NOT on campus and I live about 15 miles away from campus so it's hard to get out there

Answer 44

well that sucks

Answer 45

do you want me teach you how to solve
\[y'+p(x)y=q(x)\]

Answer 46

If i had time to learn anything more than doing this home work I would definitely be up for it, but this stuff is due tomorrow unfortunately

Answer 47

ok good luck i hoped i helped a little

Answer 48

Lol well I'm sure i'll be posting more on open study throughout this semester so keep an eye out for me! =)

Answer 49

Hell... I'm probably gonna post more today to be honest