Question

observing linearity in differential equations..!!
any comment would be appreciated..

Answer 1

@hartnn ..could you please help me out..?

Answer 2

@ganeshie8 , @dan815 ,@radar ..please help me out..:(

Answer 3

@radar @rajathsbhat @Ryaan @divu.mkr @Compassionate @texaschic101 @demitris @experimentX @UnkleRhaukus @TuringTest ..anyone please help..!!

Answer 4

basically y2 is meant to be a solution to homogeneous equation i.e the same equation which we have used to find out the solution as y1 but R.H.S equated to zero
sources: MIT 6.02X circuit and electronics math review

Answer 5

ok i understood what you meant here that why won't we take both solutions as non homogeneous one..i will try to explain as i thought should be the reason
if \[f(x_1,y_1)=4\]
and \[f(x_2,y_2)=4\]
then adding them to prove linearity would fail on the very first step because \[f(x_1,y_1)+f(x_2,y_2)=4+4=8\]
so now the RHS has turned into 8 instead of 4
but if we do like this with one y2 homogeneous solution
if \[f(x_1,y_1)=4\]
and \[f(x_2,y_2)=0\] y2 being homogeneous one
the we have\[f(x_1,y_1)+f(x_2,y_2)=4+0=4\]
now the RHS is maintained to 4 hence we have successfully made upto first step
but yeah taking both solution as non homogeneous could be the case when RHS is already 0

Answer 6

this was what explained in both my text book and video lectures..i could not think of another way round..!!

Answer 7

So, I don't know, Sorry for being helpless

Answer 8

nevermind..its really great help that you commented..atleast you put forward your opinion thanks..!!

Answer 9

f(x,y) = y' + x^2 - 4 i believe is affine due to the constant

for example:

f(x) = mx + b; f(y) = my + b

f(x) + f(y) = m(x+y) + 2b [\(\ne\)] f(x+y) = m(x+y) +b

Answer 10

but thats just a hunch ....

Answer 11

so we can't prove linearity by this method then..?
we have to look for another method then..!

Answer 12

homogenous solutions, when the constant is 0, are linear. I think its the constant that throws a monkey wrench into the works.

which is why we tend to solve the =0 first then adapt that with variation of parameters and such

Answer 13

\[y=\sum_0 c_nx^n\]
\[y'=\sum_1 c_n~nx^{n-1}\]
\[x^2+\sum_1 c_n~nx^{n-1}=4\]
\[\sum_1 c_n~nx^{n-1}=4-x^2\]
\[c_1+\sum_2 c_n~nx^{n-1}=4-x^2\]
\[c_1+2c_2x^2+\sum_3 c_n~nx^{n-1}=4-x^2\]
comparing coefficients ....
\[c_1=4~:~c_2=-\frac 12\]
tend to forget how to approach that from there ....

Answer 14

y = c + 4x - x^3/3 is the general solution :)

Answer 15

that's really cool.. general approach on finding any solution for differential equation..!!

Answer 16

but still having doubts on proving linearity on basis of superposition principle..

Answer 17

refresh me on what the superposition principle is

Answer 18

if y1 and y2 are solutions then so is a linear combination

Answer 19

yes..

Answer 20

and are you accounting for the arbitrary constants?

Answer 21

in general, if we can determine the homogenous solution: y = yh = 0

we can use that to address the nonhomogenous solution such that:

yp = A(x) yh
y'p = A'(x) yh + A(x) y'h
etc...

springs to mind

Answer 22

i believe that what the 4+0 is relating to in your previous comments

Answer 23

y'h+x^2 = 0

y'h = -x^2

yh = -x^3/3 + C ; let C be some arbitrary function of x; f(x) and solve for a particular solution yp

yp = -x^3/3 + f(x)

y'p = -x2 + f'(x)

insert into the orginal equation

(-x2 + f'(x)) + x^2 = 4

-x2 + f'(x) + x^2 = 4

f'(x) = 4, f = 4x + C

now by superpositions: y = yh + yp

something like that i beleive

Answer 24

the y exactly proves the linearity by taking y=yh+yp
it will definitely verify the superposition principle but will x follow along it ..i mean we can always put y=yp+yh to check for the linearity but can we verify along by taking x=xh+xp with y=yp+yh..will x also follow this symmetry ..?
in the previous comment you have taken everywhere same x
but if i start like this
\[y_h \prime = -x_h^2\]
\[y_p \prime=-x_p^2+f(x) \prime\]
taking x as \[x_p for y_p\] and \[x_h for y_h\]
then we can definitely superpostion y as=\[y_h+y_p\] but can we also superpostion x as=\[x_h+x_p\]
won't this leave the additional factor of \[2.x_h.x_p\]
in \[x^2=(x_p+x_h)^2=x_p^2+x_h^2+2.x_p.x_h\]

Answer 25

though many thanks for your effort..kudos..!!

Answer 26

im not that adept to determine the validity of all that :)

Answer 27

what do u want help on?

Answer 28

how super position principle works?

Answer 29

when the effect of 2 impulses on a system can be measured by taking the measurement of each each individual impulse

Answer 30

that is the super position principle

Answer 31

you will see it most commonly in the effects of Gravitational Feilds, and Electric Feilds

Answer 32

Electric Potentials