Question

Solve the given differential equation:

Answer 1

\[yy''+(y')^2=0\]

I think I'm suppose to use some sort of v-sub or using variation of parameters to solve it, but I'm not exactly sure how.

If v=y', then v'=y''. But what is y?

Answer 2

I think I have to change the independent variable? Right now it's in terms of t, but once I make the v-sub then I could make it in terms of y, so y becomes the independent?

Answer 3

\[(y' y)'=y'' y+y' y' =y''y+(y')^2\]

Answer 4

\[y y''+(y')^2=0 \\ (y' y)'=0\]

Answer 5

Hmm, but what about in the case of
y''+y=0 or y''+y(y')^3=0?

Answer 6

I don't necessarily need those answers, but I think there's a different approach besides inspection to see that the combination is the product rule =/ Because that's not always the case

Answer 7

I know it isn't always the case but it worked here :p

Answer 8

y'' + y = 0
is a regular linear equation which you can solve by using characteristic eqn thingy

Answer 9

Okay, but it wants me to solve using the substitution technique XD I should've made that more clear. I just wasn't sure how to go about doing that exactly

Answer 10

use the sub u=y' y
:p

Answer 11

\[u=y' y \\ u'=y''y +(y')^2 \\ \text{ so we have } u'=0\]

Answer 12

That's still by inspection though :P

I managed to find a similar solution, and their approach was that
\[v=y', v'=y'', \frac{ dv }{ dt }=v \frac{ dy }{ dt }\]
But I don't exactly understand this either..

Answer 13

also please don't get mad at me
I used to have blonde hair

Answer 14

In each case, you may let \(\dfrac{dy}{dx}=v\) and get
\[y''=v' = \dfrac{dv}{dx} = \dfrac{dv}{dy}\dfrac{dy}{dx} = \dfrac{dv}{dy}v\]

Answer 15

notice that we're assuming \(v\) as a function of \(y\), not \(x\)

Answer 16

we're actually gettign rid of the independent variable \(x\),
that allows us to work with the differential equaiton with indepent variable \(y\) and dependent variable \(v\)

Answer 17

@freckles lol I'm not mad :P

Okay, that makes a little bit more sense!

Answer 18

for example,
\(y''+y=0\)

changes to the system of first order eqns
\(\dfrac{dy}{dx}= v(y)\)

\(\dfrac{dv}{dy}v(y)+y=0\)

Answer 19

as you can see, the equation is trivial to solve now because it is separable :
\(\dfrac{dv}{dy}v(y)+y=0\)

\(\int v(y) dv = \int -y dy\)

Answer 20

once you have solved \(v(y)\), you can solve \(y(x)\) using the other eqn : \(\dfrac{dy}{dx}=v(y)\)

Answer 21

Ahh, that makes sense now. Thanks!!

Answer 22

np