Question

y"+αy'+βy = 0. let sum of the roots of its auxiliary eq be 94 , product of the roots is 2013. how to solve the DE.

Answer 1

should we assume an e^rx solution, or some x^r solution?

Answer 2

the key might be in getting a system of equations from roots

Answer 3

r^2 + ar + b = 0 when -a +- sqrt(a^2 -4b) = 0

Answer 4

should we use the sum n the product of the roots ?

Answer 5

yes, but im not too sure if we need to run thru the auxilary stuff

r1 + r2 = 94
r1 * r2 = 2013

should develop roots that fit the bill

Answer 6

based on sum = 94 and product = 2013
we can conclude that
a = 94
b = 2013

Answer 7

(47-n) +(47+n) = 94

47^2 - n^2 = 2013

solve for n :)

Answer 8

it looks like n = 14

47+14 = 61
47-14 = 33

Answer 9

i wonder if these are suitable roots ....

Answer 10

okay.. but if i use the info from the question given, \[\alpha + \beta = 94 , \alpha \beta = 2013\] how to get the value of alpha and beta ?

Answer 11

alpha and beta would be a part of the auxil eq such that\[r=\frac{-a\pm\sqrt{a^2-4b}}{2}=61,33\]

Answer 12

-a/2 = middle of 61 and 33; 47

a = -94

Answer 13

\[\frac{\sqrt{94^2-4b}}{2}=14\]
\[\sqrt{94^2-4b}=28\]
\[94^2-4b=28^2\]
\[\frac{94^2-28^2}{4}=b=2013\]
just like dumbcow prophecised :0

Answer 14

I've seen this exact question here before: http://openstudy.com/study#/updates/51a0f70ae4b0e9c70c332ec0

@amistre64 and @dumbcow have got it covered, though.

Answer 15

okay..thanks for your help @amistre64 @dumbcow @SithsAndGiggles ;)