Question

Constant multiples of solutions question....

Answer 1

Show that for any constant C, the function \[Ce ^{-x}\]is a solution of equation \[\frac{ dy }{ dx }+y=0\] while \[Cx ^{-1}\]is a solution of equation \[\frac{ dy }{ dx }+y ^{2}=0\]only when C=0 or 1

Answer 2

\[(Ce ^{-x})' = -Ce ^{-x} \\(Ce ^{-x})'+Ce ^{-x} =0\]
let set \(Ce ^{-x} =y\) so, it becomes y' +y =0
or \(\dfrac{dy}{dx}+y =0\) DAT SIT

Answer 3

for the second part, just put c =0 or c =1 to find the solution

Answer 4

I think I've got it

Answer 5

I know!! sometimes, we are panicked and turn blank to a very simple problem. :) me too!!!

Answer 6

How about this one:

Show that for any linear equation of the form \[\frac{ dy }{ dx }+P(x)y=0\] if y(x) is a solution, then for any constant C the function Cy(x) is also a solution.

Answer 7

to me, I would like to solve it
to get the form of y . what do you get the form of y in term of P(x) ?

Answer 8

hmmmm give me a min

Answer 9

oh, no need to solve!! I am sorry. just plug back

Answer 10

do you get the integrating factor to be \[e ^{\int\limits_{?}^{?}P(x)}\]

Answer 11

if y( x) is solution, then y'(x) +P(x) y(x) =0
try Cy(x) , (C(y(x))' = C y' , then Cy' + P Cy = 0 or not? surely it is , just factor C out,
C [ y'(x) + P(x) y(x) ]=0 (from the condition above) --> Cy(x) is a solution, too

Answer 12

is it unclear?

Answer 13

I understand, it seems much more simple than what I was trying to do.

Answer 14

we have the condition that : y'(x) + P(x) y(x) =0, right?
now, try Cy(x) , if it is a solution, then it satisfies the condition which says
(solution)' + P(x) *(solution) = 0

Answer 15

hahaha that is clear as mud

Answer 16

hahahaha... mud is clean or clear???

Answer 17

just kidding.....I understand it....lol

Answer 18

mud is neither clean or clear but it may be sterile

Answer 19

are you jonnymiller or ponnymiller ??(joke)

Answer 20

so for this problem it seems that for any given linear diff eq the solution will be a specific function and that function can have any constant multiple. for example: y=x^2 could be a solution or y=2x^2 or y=3x^2.

Is my thinking correct?

Answer 21

just for this problem only, (the first order ), to the second order, things are different!!

Answer 22

yes I meant to say for first order linear diff eq's

Answer 23

Thanks for your help!