Calculus: how do I solve this initial value problem?

Question

Using the fact that h(x)=ln{x+SQRT(1+x^2) has derivative h'(x) =1/{SQRT(1+x^2)} solve the initial-value problem dy/dx=y/{SQRT(1+x^2) ,
(y greater than 0 ), y = 2 when x = 0, giving the solution in explicit form.

Right, I have taken the y accross to the LHS and then integrated both sides and I got y^2=2ln{x+SQRT(1+x^2) + C where C is a contstant. I think that is ok but I don't know what to do from there !!

Answer 1

First, I don't think your solution so far is correct. Rearrange the differentials in your problem:

(1/y) dy = 1/sqrt(1+x^2) dx

Integrating both sides, you get:

ln(y) = ln( x + sqrt(1+x^2) ) + D

Where "D" is some constant. Note that I can pick a different constant C such that:

D = ln(C)

In that case, I get:

ln(y) = ln( C( x + sqrt(1+x^2) ) )

Now, I can remove the "ln" from both sides to give me:

y = C*(x + sqrt(1+x^2) )

Now, to check, take the derivative of that:

dy/dx = C*( 1 + x/sqrt(1+x^2) ) = C*(x + sqrt(1+x^2))/sqrt(1+x^2)

This is correct. dy/dx = y/sqrt(1+x^2).

Now, to solve for "C", use the conditions you were given for (x,y)=(0,2). Just plug in for x and y and solve for C:

2 = C*(0 + sqrt(1+0^2) ) = C*sqrt(1) = C

So C=2.

So your solution is:

y = 2*( x + sqrt(1+x^2) )

Answer 2

Now put y=2 and x=0 in y^2=2ln{x+SQRT(1+x^2) + C so that

4= 2 ln {0+sqrt (1+0)} +C
4= 2 ln (1) + C

C= 4 (as ln 1 = 0)

Therefore y^2=2ln{x+SQRT(1+x^2) + 4