Question

Differential equations:

*please see attached pic under comments

Answer 1

reduction order is finding another solution whereby:

y2 = v y1

Answer 2

or the formula that comes about from it:
\[y_2=y_1\int \frac{e^{-\int P(x)}}{(y_1) ^2}\]

Answer 3

i'm supposed to find a way to reduce it to first order is it?

Answer 4

yep

Answer 5

we know:

y2 = v y1

y2 = v t^-1

y2' = v' t^-1 + -v t^-2

y2'' = v'' t^-1 + -v' t^-2 + -v' t^-2 + 2v t^-3

Answer 6

simplify and insert into your "equation" and it reduces the order

Answer 7

type up the equation in here so i dont have to keep going back and forth ....

Answer 8

..cant even copy and paste it in that format ..

Answer 9

okay lol sorry. just a minute.

Answer 10

The equation is:

\[t ^{2}y"-3ty'-5y=0,t >0\]

Answer 11

thats better :)

Answer 12

now lets plug in our founded v primes

Answer 13

t^2(v'' t^-1 -2v' t^-2+ 2v t^-3)

-3(v' t^-1 + -v t^-2)

-5(v t^-1)

does this make sense so far?

Answer 14

we basically plugged in the v' that we got from what we got when we used the formula that you first mentioned right?

Answer 15

i meant the reduction order..first

Answer 16

yes, all the related parts for y'' y' and y; just substitute them into the diffy q and expand/simplify


v'' t -2v'+ 2v t^-1

-3v' + 3v t^-1

-5v t^-1



v'' t + (-2-3)v'+ (2+3-5)t^-1 v



v'' t -5v'+ 0t^-1 v = 0

chk the math to make sure i didnt forget how to add

Answer 17

looks good to me, but then again, im blind :)

now that its in this for we simply rename the v' as say ... m

v'=m
v'' = m'

m' t -5m = 0

is this more doable

Answer 18

once we know "m" then:

\[\int (v' = m)\implies v=\int m \]

Answer 19

yes i think i's all been good so far.. for this one you just did derivatives right?

v'' t -2v'+ 2v t^-1

-3v' + 3v t^-1

-5v t^-1

Answer 20

yes

y(x)2 = v(x)*y(x)1

then its just derivatives using the product rule as a guide

Answer 21

since we need to know y, y' and y'' we need to know what to substitute in

Answer 22

ugh, im writing xs for ts ... hope thats not confusing you

Answer 23

\[m' t -5m = 0\]
\[m' -5/t\ m = 0\]
\[e^{ln(t^{-5})} m = \int 0\]
\[e^{ln(t^{-5})} m = C\]
\[m = Ct^5\]
\[\int (v' =m) = \int Ct^5\]
\[v = \frac{Ct^6}{6}+c_1 \]
\[y_2= vy_1:\ \frac{Ct^6t^{-1}}{6}+c_1t^{-1} \]
\[y_2= t^5\]

Answer 24

we can dbl chk that with the formula:
\[y_2 = t^{-1}\int \frac{e^{-\int -3/t}}{t^{-2}}\]
\[y_2 = t^{-1}\int \frac{t^3}{t^{-2}}\]
\[y_2 = t^{-1}\int t^5\]
\[y_2 = t^{-1}(\frac{t^6}{6}+C)\]
\[y_2 = t^5\]

Answer 25

it's fine I did understand that the x's were meant to be t's.. this is the best explanation that I have ever encountered with diff eq problems so far.. It's really understandable. I wish you could just be my prof.. lol..jk.. Thanks a lot! I appreciate you going through the trouble to try to explain this to me.. Thanks! You were awesome..

Answer 26

youre welcome, it helps that we just went over this in class last week ;)