Question

Just a hint please. Solve: \[y'=\frac{t^2+y^2}{ty}\]

Answer 1

I don't think I can use separable variables. I cant see integrating factor...

Answer 2

whats your question? sorry

Answer 3

o

Answer 4

\[y'=\frac{t^2+y^2}{ty}\]

Answer 5

Solving the question stated

Answer 6

but I only want a hint

Answer 7

are u trying to fidn the anti derivative?

Answer 8

no, it's an ODE

Answer 9

I'm solving for y.

Answer 10

split the fraction ...

Answer 11

i just did it i dont understand it either

Answer 12

I did that, but it left me with nothing helpful

Answer 13

\[y'=\frac{t}{y}+\frac{y}{t}\]

Answer 14

try letting y/t = v, therefore vt=y

Answer 15

k give me a minute

Answer 16

so \(v'=v^{-1}+v\)?

Answer 17

so then \[v'=\frac{1+v^2}{v}\]

Answer 18

let me chk

y' = v't + v

v't + v = v^(-1) + v

Answer 19

then I can use separable

Answer 20

brilliant!!! Thank you!!!!

Answer 21

\[v't+v=v^{-1}+v\]
\[v't=v^{-1}\]
\[v ~dv=\frac1t dt\]
is what im getting, or do we see an error in it?

Answer 22

I didn't get that

Answer 23

if y = vy

what is y'?

Answer 24

so y=vt y'=v'

Answer 25

vt is a prduct

Answer 26

yep, but we are differentiating wrt t no?

Answer 27

yes, but its still a product of 2 function related by t

Answer 28

awe phooey

Answer 29

x f(x)

how wold we diffrentiate this?

Answer 30

give me a minute to fix it

Answer 31

now I get the same as you

Answer 32

\(vdv=\frac{1}{t} dt\)

Answer 33

I got it from here, will you check my final result?

Answer 34

sure, but i usually mess it up from here :)

Answer 35

You can check to see if the ODE is homogeneous using the following sort of procedure. \(y'=f(y,t)\) is homogeneous if you can write \(f(\lambda y,\lambda t)=\lambda^k f(y,t)\) for some real \(k\). If you can establish this, the substitution amistre suggested will always work.

Answer 36

ok so I got,\[ y=t\sqrt{2ln|t|+c_1}\]

Answer 37

v^2 = ln|t| + c

v = sqrt(ln|t| + c)

y/t = sqrt(ln|t| + c)

y = t sqrt(ln|t| + c)

kinda wonder where you got that 2ln(t) from

Answer 38

integral of v is 1/2 v^2

Answer 39

thnx sith, these things work by way of repressed memorys for me :)

Answer 40

doh .... and 2ln|t| = ln(t^2) if you feel like it

Answer 41

yea, but it's unnecessary since i'm not doing any e^ln stuff

Answer 42

Thank you! I appreciate your help!

Answer 43

your welcome ... srry i forgot how to integrate v :) knew id mess it up on this end somehow

Answer 44

haha, you got the important part though!