Question

how to solve the following differential equation ??

Answer 1

\[\frac{d^2q}{dt^2}+\omega^2q=0\]

Answer 2

@ganeshie8

Answer 3

please try this substitution:
\[q\left( t \right) = c\exp \left( {\lambda t} \right)\]

Answer 4

you should get a quadratic equation for lambda

Answer 5

of course, c and lambda are two constants

Answer 6

\[\frac{d}{dt}ce^{\lambda t}=\lambda ce^{\lambda t}\]\[\frac{d^2}{dt^2}ce^{\lambda t}=\frac{d}{dt}\lambda ce^{\lambda t}=\lambda^2ce^{\lambda t}\]\[\lambda^2ce^{\lambda t}+\omega^2ce^{\lambda t}=0\]

Answer 7

ok! so we have to solve this equation:
\[{\lambda ^2} + {\omega ^2} = 0\]

Answer 8

solve for lambda or omega?

Answer 9

sorry this site lagged out,
I solved for lambda and i got
\[\lambda=\pm j \omega, j=\sqrt{-1}\]

Answer 10

@shamim

Answer 11

@phi

Answer 12

ok. now test the solution in the original equation

Answer 13

in the differential equation?

Answer 14

yes

Answer 15

\[\lambda^2e^{\lambda t}+\omega^2e^{\lambda t}=-\omega^2e^{j \omega t}+\omega ^2e^{j \omega t}=0\]

\[q=e^{jwt}\] is the solution?

Answer 16

one of the solutions. also -jwt for the exponent.
also, don't leave out the constant of integration c

Answer 17

\[q=Ce^{\pm jwt}\]?

Answer 18

yes

Answer 19

that's it?

Answer 20

the answer would be a linear combination of the two solutions
\[ q = C_1~ e^{j \omega t} + C_2~ e^{-j \omega t} \]

Answer 21

why is that?

Answer 22

we know the first expression (with jwt) results in 0 when we put it into the diff. eq.
also, the second expression results in 0 (you can check this)
if both expressions result in zero, the sum of both expressions will also be zero
i.e. a solution of q'' + w^2 q = 0

Answer 23

ahh, right 0+0=0 right?

Answer 24

the linear combination is the most general solution, and depending on C1 and C2, we can generate any specific solution (which is determined by initial conditions, if we know them)

Answer 25

but this is a solution with imaginary unit, is there no real solution?

Answer 26

it depends in the unknown coefficients.
e^jwt = cos(wt) + j sin(wt)
and
e^-jwt = cos(wt) - j sin(wt)

if C1= C2= 1
then we could get the real solution 2 cos(wt)

Answer 27

in my book solution is given
\[q=q_{0}\cos(\omega t)\]
\[q_{0}\] is constant, how can I get to there from this ?

Answer 28

also what formula is that ??

Answer 29

which formula? can you be more clear ?

Answer 30

e^jwt=coswt+jsinwt

Answer 31

Euler's formula
see https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/maclaurin_taylor/v/euler-s-formula-and-euler-s-identity
or
https://en.wikipedia.org/wiki/Euler%27s_formula

the book's solution assumes the answer is real. if we are told we only want real solutions, then
\[ q = C_1~ e^{j \omega t} + C_2~ e^{-j \omega t} \\
q= (C_1 + C_2 ) \cos \omega t + j \left( C_1-C_2\right) \sin \omega t
\]

Answer 32

to make q real, you must set C1= C2 (which causes the imaginary part to be 0)
thus a pure real solution would be
\[ q = 2C \cos \omega t = q_0 \cos \omega t \]
where C1+C2 = 2C is renamed \(q_0\)

Answer 33

\[C_{1}=C_{2}=\frac{q_{o}}{2}\]

Answer 34

remember C1 and C2 are arbitrary
if you add two arbitrary numbers or divide/multiply by a constant factor, you get another number. But because it's arbitrary, you can call it a new name, and pick it directly
Does that make sense?

Answer 35

Yea I understand that
Ok I looked it up and I found that it can be shown for this equation that \[C_{1}\] and \[C_{2}\] are complex conjugates, what does that mean?

Answer 36

example: C+1
if C is arbitrary we can make the expression *any number*

on the other hand, if we rename C+1 as D (another arbitrary value)
we can pick D to be any number.

the only time we have to be careful is if we had (for example)
C* exp(jwt) + (C+1)* exp(jwt)
we can't rename C+1 to D (without renaming the first term to (D-1) )
because the constants are related.

Answer 37

Yeah, that makes sense Isn't that what you do in Integrals anyway, add up the constants?

Answer 38

a+ j b
a - j b
are complex conjugates

if x and x* are complex conjugates then
x x* = |x|^2

Answer 39

alright then,
\[q=(C_{1}+C_{2})\cos(\omega t)+j(C_{1}-C_{2})\sin(\omega t)=2acos(\omega t)-2bsin(\omega t)\]

Answer 40

ok I guess this is rapped up then, thanks!

Answer 41

yes, if we say C1= a+j b and C2= a - j b
now, because a and b are arbitrary, then so is 2a and -2b
so rename to q1 and q2:
\[ q = q_1 \cos \omega t + q_2 \sin \omega t \]

to summarize, to get a real solution we require the C1 = conj(C2) (which is more general than what I original thought, C1=C2)