Question

How do I differentiate v''t^2=0

Answer 1

what is that
\[v"t^2=0\]
differentiate with respect to what?

Answer 2

The original question was this

Answer 3

I learned second order ODE when i was in high school, but they dont teach reduction of order, so I am gonna take a guess

Answer 4

Clearly, t is solvable from the given, so solve for t first. then plug back and solve for y. my guess is that you will get two values for y. One is the given t^2, the other will be your answer

Answer 5

that's just my guess so no guarantee it works or it is right

Answer 6

you mean how to solve for t or how to solve for y?

Answer 7

the solution for y2

Answer 8

https://www.youtube.com/watch?v=oQSFW8BIrY0

Answer 9

in his example he uses v'' and v' in mine I have v'' and v and I'm not sure how to equal w to that

Answer 10

t^2y'' -4ty' +6y = 0, y1 = t^2

let

y2 = vt^2
y2' = v't^2 + 2vt
y2'' = v''t^2 +4v't +2v

plug them in ...

t^2(t^2 v'' +4t v' +2v) -4t(t^2v' + 2t v) +6t^2 v = 0

t^4 v'' +4t^3 v' +2t^2 v -4t^3 v' -8t^2 v +6t^2 v = 0

combine like terms

t^4 v'' +(4t^3 -4t^3) v' (-8t^2 +6t^2 +2t^2) v= 0

Answer 11

so, t^4v'' = 0 is what i see this as

Answer 12

yeah i was overthinking it when I redid it lol, but how do I continue from there

Answer 13

solve for v of course

Answer 14

坑爹啊！！！linear methods don't work on non-linear equations!!

Answer 15

That's what I mean I don't know how to solve that.

Answer 16

this is a linear equation ...

Answer 17

OK, I guess I won't get banned using Chinese

Answer 18

no, cuz they don't have constant coeffs

Answer 19

lol, its all diamonds to me

Answer 20

at least that's what my high school math teacher said

Answer 21

this is a 'second-order' _linear_ ordinary differential equation

Answer 22

are my vs primed correctly? chk the algebra to make sure its good to this point

Answer 23

yeah it's correct I got that too

Answer 24

I still don't know how the answer is t^3 :l

Answer 25

reduce

let w = v'
w' = v''

t^4 w' = 0

let z = w'

t^4z = 0

but thats seems redundant to me, but we can integrate both sides

Answer 26

forget the z

Answer 27

divide off the t^4

w' = 0

integrate, w = C for some arbitrary constant

Answer 28

w = v' = C

v' = C, integrate again

v = t + c

Answer 29

y2 = vt^2

y2 = (t+c)t^2

y2 = t^3+ct^2, drop the y1 parts of it

y2 = t^3

Answer 30

gottt it you're a life saver

Answer 31

recall w' means dw/dt

dw = 0 dt, and constants derive to 0

w = some constant ... etc

Answer 32

yeah totally missed that when I was doing it thank youuu

Answer 33

good luck :)