? solution of ( dy / dx ) + ( y / (sqrt(a+(x^2))) ) = 0

If y = f(x) , and ( dy / dx ) + ( y / ( sqrt(a+(x^2)) ) ) = 0

I knew its solution is y = { sqrt(a+(x^2)) - x } , where a is a constant

can any one give the proof , by solving the differntial equation.

Are there any other solutions for the above given differential equation. I asked this other
solutions because, on rearranging the given differential equation we get

( dy / y ) = - ( dx / ( sqrt(a+(x^2)) ) )
on integrating
ln y = { integral ( - ( dx / ( sqrt(a+(x^2)) ) ) ) } + ln c

so I may get the solution as y = ce^y1 , where y1 is funcion of x other than f(x)

I am not sure about the existance of general solution , but I think it may exist.

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so what I want is ,

1. Solving procedure for the differntial equation to get the solution
y = { sqrt(a+(x^2)) - x }

2. what is general answer for { integral ( - ( dx / ( sqrt(a+(x^2)) ) ) }

3. Is there any general solution for the differential equation given.
other than y = { sqrt(a+(x^2)) - x }

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Procedures I have tried and falied to do further.

1. we know d(sqrt(a+(x^2))) / dx = x / sqrt(a+(x^2))
so 1 / sqrt(a+(x^2)) = ( d(sqrt(a+(x^2))) / dx ) /x
on substituting this value in the differential equation , we will get
( dy / dx ) + ( ( y * d( sqrt(a+(x^2)))/dx ) / x ) = 0
on solving this I got strucked at
ln y = ( - ( sqrt(a+(x^2)) ) / x ) - { integral ( sqrt(a+(x^2)) / x^2) dx )

2. rearranging the differntial equation we get
( dy / y ) = - ( dx / ( sqrt(a+(x^2)) ) )
take x = a cos(t)
dx = - a sin(t) dt
t = cos^-1 (x/a)

on solving I got strucked at
( dy / y ) = { ( sqrt(2p) * sin(t) ) / sqrt( cos(2t) + 3 ) } dt

I got no other ideas. I wish , I can get the answers for all the 3 questions I have asked.

I think you began correctly separating the variables, that is:

ln = - integral [ ( dx / sqrt(a+x^2) ]

now let t = x/sqrt(a); "a" must be positive, then: ln = - integral [ dt / sqrt(1+t^2) ]

remembering that d/dt[argsinh(t)] = 1/sqrt(1+t^2) and that argsinh(t) = ln[t + sqrt(1+t^2)] you will obtain:

ln = -ln [t + sqrt(1+t^2)] ==> y = 1/[t + sqrt(1+t^2)] multiplying now by [t - sqrt(1+t^2)]/[t - sqrt(1+t^2)]:

y = [sqrt(1+t^2) - t] but t = x/sqrt(a) from which:

y = [sqrt(1+x^2/a) - x/sqrt(a)] = 1/sqrt(a) * [sqrt(a+x^2) - x]

Of course there will be also an additive constant (omitted).

I'm not sure if the coefficient "1/sqrt(a)" is my mistake I have to check better the calculations, however the procedure should be OK.

albbg said:

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(1) ln = -ln [t + sqrt(1+t^2)] ==> y = 1/[t + sqrt(1+t^2)] multiplying now by [t - sqrt(1+t^2)]/[t - sqrt(1+t^2)]:

y = [sqrt(1+t^2) - t] but t = x/sqrt(a) from which:

(2) y = [sqrt(1+x^2/a) - x/sqrt(a)] = 1/sqrt(a) * [sqrt(a+x^2) - x]

Of course there will be also an additive constant (omitted).

Click to expand...

The additive constant is at step (1)
At step (2) becomes multiplicative and includes the term 1/sqrt(a)
--> y = C*[sqrt(a+x^2) - x]

_Eduardo_, you are right. Thank you.