IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

subs

Can any one tell me how the solution to
x''+ kx = 0(changed it!!)

is Asin(wt+-d) and Acos(wt+-d)

rebel

~~~~~~~~~~is it double dash x ????~~~~~~~~~~~~~~~~~~~~~~~~

subs

yahiyafirdous

Mr subs
Try to present your question correctly, you have made following mistakes while framing your question.
(1) f''(x) means d²f(x)/dx^2,but observing your answer it appears , you mean f''(x)=d²x/dt².
(2) you have not mentioned anything about w, it is not an arbitrary constant. It shouls come through solution of the equation.

Now I am presenting you a very simple method to solve this.

SOLUTION:
d²x/dt²= -kx
Let dx/dt=p
Hence d²x/dt²=(dp/dx)(dx/dt)=p(dp/dx)

p(dp/dx)= -kx

pdp= -k

xdx
p²= -kx² + C₁ where C₁ is an arbitrary constant of integration.
dx/dt=

[k(C₁/k - x²)

dx/

[k(C₁/k - x²) =

dt
sin^-1(kx/C₁) = (

k)t + C₂ where C₂ is another constant of integ.
x=(C₁/k)sin(t

k + C₂)

Here let A=C₁/k,

k=w, d=C₂

x=Asin(wt+d)

Whether d is positive or negative, it is determined by conditions given in problems.

avinash.sharma

yahiyafirdous has made reasonable comments on the question.

We can write the question as

d²x/dt²+kx =0

or

(D²+k)x = 0 where D= d/dt

then auxiliary equation is m²+k = 0 or m=

(-k) = i

k

and so C.F. (complementary function) = C₁ cos {

k .t} + C₂ sin {

k . t} where C₁ and C₂ are arbitrary constants.

C₁ and C₂ can be resolved by other conditions.

dvrravi

aux eqns are out of syllabus for jee.
we only have to use variable saperable method and such for a first orfer diff eqn.
this question is void.it won't be asked..

subs

it maybe void in maths... however this is what SHM in phy is.....

so jus wanted to clarify how we get this!

dvrravi

substitute the values in the eqn....

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