Question

Form a suitable differential eqn using
y=Ax^2+Bx^5

Answer 1

\(y'=2Ax+5Bx^4\)
\(y''=2A+20Bx^3\)

Answer 2

can I divide
\(\frac{y'}{y''}\)
?

Answer 3

yolo if you still need help lmk i can do it

Answer 4

and yeah you can, it depends on what kind of differential equation they want. if you divide i don't think it's linear.

Answer 5

They didnt specify what kind of differential eqn.
Yes.I do need help XD

Answer 6

yep,I trying to figure out how to eliminate A and B

Answer 7

this equation is the solution to a class of differential equations called the euler I believe

Answer 8

http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx

Answer 9

if you're familiar with differential equations it should be enough for you to come up with the equation that you need, but if not i can probably look further

Answer 10

Not really familiar XD
but thanks for the info
Is there any other way besides using euler eqns

Answer 11

depends on if you need to remove A and B. But honestly just reading the first few lines of that should be enouhg

Answer 12

The question ask to eliminate A and B.
But do u know how to eliminate those constants?

Answer 13

yes, just formulate the appropriate euler differential equation by working backwards from the link I gave you

Answer 14

aite,might take me a while to digest it.

Answer 15

well there is an easier way I guess

Answer 16

y=Ax^2+Bx^5
y'=2Ax+5Bx^4
y''=2A+20Bx^3
y'''=60Bx^2
y4=120Bx
y5=120B (y5=y''''')

y''=2A+20Bx^3
y''=2A+20(y'''''/120)x^3
A=y''/2-y'''''/12*x^3

y=(y''/2-y'''''/12*x^3)*x^2+y'''''/120*x^5

y=y''/2*x^2-y'''''/12*x^5+y'''''/120*x^5
y=y''/2*x^2-(9/120)*y'''''*x^5

Answer 17

granted i don't think that was easier cause that took more thinking than solving the euler differential equation did

Answer 18

lol,yes XD
I think I will just use euler eqns
If I got any other solutions,I share it here ^

Answer 19

Thanks bro ^

Answer 20

yup no problem

Answer 21

We cant differentiate y up to 5 times because there are only 2 arbitrary constants
Since we were given 2 arbitrary constants in the solution , the original diff. eqn. will hv at most 2 derivative-that the most derivatives

Answer 22

\((1) \quad y = Ax^2 + B x^5 \\ \, \\ (2) \quad y' = 2A x + 5Bx^4 \\ \, \\ (3) \quad y'' = 2A + 20B x^3\)

Multiply equation (2) by x, and equation(3) by x^2

\((1) \quad y = Ax^2 + B x^5 \\ \, \\ (2) \quad xy' = 2A x^2 + 5Bx^5 \\ \, \\ (3) \quad x^2 y'' = 2Ax^2 + 20B x^5\)

Multiply equation (1) by 2

\((1) \quad 2y = 2Ax^2 + 2B x^5 \\ \, \\ (2) \quad xy' = 2A x^2 + 5Bx^5 \\ \, \\ (3) \quad x^2 y'' = 2Ax^2 + 20B x^5\)

Subtract, equation (2) - (1)
and equation (3) - (1)

\((1) \quad 2y = 2Ax^2 + 2B x^5 \\ \, \\ (2) \quad xy' - 2y = 0 + 3B x^5 \\ \, \\ (3) \quad x^2 y'' - 2y = 0 + 18 Bx^5\)

multiply equation(2) by 6

\((1) \quad 2y = 2Ax^2 + 2B x^5 \\ \, \\ (2) \quad 6xy' -12y = 0 + 18B x^5 \\ \, \\ (3) \quad x^2 y'' - 2y = 0 + 18 Bx^5\)

Subtract equation(3) - equation(2)

\((1) \quad 2y = 2Ax^2 + 2B x^5 \\ \, \\ (2) \quad 6xy' -12y = 0 + 18B x^5 \\ \, \\ (3) \quad x^2 y'' - 2y - 6xy' + 12y = 0 + 0\)

Then equation (3) gives us
\(x^2 y'' -6xy' + 10y = 0\)

Answer 23

that's a good point that i didn't realize @Logic007 . Glad you figured it out lol

Answer 24

yeah,I just realized it XD
I got another question related to separable method

Answer 25

those are fun and easy... is this for an ordinary differential equations class?

Answer 26

yep,that's right
I just finish sem 1 finals
rn,it is sem break for me
But I decided to to finish differential eqn b4 I enter sem 2 XD

Answer 27

I had a lot of fun with differential equations, though our school teaches it as applied math so we don't go in detail into the theory

Answer 28

Owh I c.
http://prntscr.com/m39l3d
This are the chapters for diff eqn subject

Answer 29

These*

Answer 30

ahh brings back memories from 3 years ago haha

Answer 31

haha XD