Question

na please help me with this one

to find the solution of the differential equation

[(x+1)^2]d^2y/dx^2 + (x+1)dy/dx + y = cos[ln(x+1)]

Answer 1

Homogeneous part:
Let t = x+1
--> \(t^2y''+ty'+y =0\) It's an Euler form.
Characteristic equation: r^2+1 =0 gives us complex root \(r=\pm i\)
so, general solution is \(y_c = C_1cos (lnt)+C_2sin (lnt)\)
replace t = x+1, we have \(y_c = C_1cos(ln (x+1) +C_2sin(ln(x+1)\)
Partial part:
y = t cos (lnt)
I think you can step up, right?

Answer 2

yep i can
but where does the euler equation form from

Answer 3

I am sorry, it's indicial equation, not characteristic equation

Answer 4

and how is r^2 + 1= 0 the complimentary part's equation giving function

Answer 5

From your original one , after let t = x+1, we get that form, just apply formula

Answer 6

r^2+1 =0 ---> r^2 =-1 ---> r = \(\pm i\)

Answer 7

yea i get that
but why will r^2 give us the general solution, where does that come from

Answer 8

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

Answer 9

That's the step to solve Euler's ODE

Answer 10

yep i get the t^2y'' + ty+ +y = 0 part is by substution, where does the r^2+1 come from

Answer 11

ahh (Y) i'll check it out

Answer 12

how/why is cos{ln(x+1)} equated to zero

Answer 13

ahahahahaha sorry na to bug you that much ^3^)7 i'll just check it out myself

Answer 14

I don't let it = 0. just separate to solve homogeneous part and partial part.

Answer 15

so the substution gives the homogenous part?

Answer 16

for convenient, you should substitute to both.

Answer 17

i shall (Y)

Answer 18

what does (Y) mean in "i shall (Y)"?

Answer 19

it means the thumbs up sign on fb XD so a bit of habit,a bad one i guess