Question

Hi everyone! How would you go about solving this:
y''+x^2y'+e^x=0 Can someone push me in the right direction? Thanks! :o)

Answer 1

find the homogenous and then the particular solutions

Answer 2

y'' + x^2 y' = 0 gives us the homogenous solution

if we multiply thru by say e^(1/3 x^3) that should help out if memory serves

Answer 3

what method is that?
I can only seem to recall how to solve constant coefficient homogeneous de's

Answer 4

what would you have multiplied thru if it was a constant there? say 3

e^(3x) right?

Answer 5

if there was a constant where exactly?

Answer 6

yousaid you remembered it, insted of x^2 there is a 3

Answer 7

I wouldn't multiply through at all...
I would make the equation r^2+3r+1=0, find the roots and apply the proper formula for the solution

Answer 8

y' + 3y = 0

let u = e^3x, u' = 3 e^(3x)

uy' + 3uy = 0

e^(3x) y' + 3 e^(3x) y = 0

this is the product rule if we integrate both sides

e^(3x)y = 1+C right?

Answer 9

yeah I don't think I have been taught this method yet...looks completely foreign to me

Answer 10

its taught before the characteristic equation method, since it relies only on what you learnt in derivatives class

Answer 11

hmm...maybe I just forgot then...so basically flip it around, integrate, and what not?

Answer 12

(uy)' = uy' + u'y

let u = e^(3x), and y=y

e^(3x) y' + 3 e^(3x)y = e^(3x) (y' + 3y)

Answer 13

yeah, determine an integration function i think is the technical name for it, that may turn it into a reverse product rule

Answer 14

okay...that's good enough...I can look back at my notes...my teacher never had us taking double integrals before though...I will review...Thanks!

Answer 15

y'' + x^2 y' = -e^x

let u = e^(??) and u' = x^2 e^(??), ?? has to be 1/3 x^3

multiply both sides by e^(1/3 x^3)

e^(1/3 x^3) y'' + x^2 e^(1/3 x^3) y' = - e^(1/3 x^3) e^x

e^(1/3 x^3) y' = -integral [e^(1/3 x^3 +x)]

y' = -integral [e^(1/3 x^3 +x)] * e^(-1/3 x^3)

Answer 16

let v = e^(1/3 x^3 +x)

ln(v) = 1/3 x^3 + x

int(ln(v)) = 1/12 x^4 + 1/2 x^2 + C

v ln(v) - v = 1/12 x^4 + 1/2 x^2 + C

there might be a simpler approach, but this is what came to mind to start with :)

Answer 17

yeah, seeing all that, I think the teacher wanted me to see if a 2nd solution besides zero existed using that big integral formula, so y=0 should be the only solution then I'm thinking

Answer 18

yeah I definitely haven't solved 2nd degree variable coeff before :o)

Answer 19

well if your satisified with it, then we are done here :) good luck and all

Answer 20

thanks for the hard work! I will always be able to come back and see you mathematical wizardry! :o)