I've integrated the following and function and the given answer is in the form of A cosh((x+a)/A). How do I simplify my answer to get this answer?

Integral
y^2/(1+((dy/dx)^2)) = c

Answer I got:

y= a*e^tx + b*e^(-tx)

Where a and b are constant and t=((c-1)/c), c is the constant above

1 answer

To simplify your answer, you need to express it in terms of hyperbolic functions. One way to do this is to use the identity:

cosh x = (e^x + e^(-x))/2

First, let's find the value of t in terms of A and a. From your solution, we have:

t = (c-1)/c = (1-1/A^2)/(1+1/A^2) = (A^2-1)/(A^2+1)

Now let's write y in terms of A and a. We have:

y = a*e^(tx) + b*e^(-tx)

= a*(e^t)^(x) + b*(e^(-t))^(x)

= a*cosh(tx) + b*sinh(tx)

= a*cosh((A^2-1)/(A^2+1) x) + b*sinh((A^2-1)/(A^2+1) x)

Now we use the identity above to express y in terms of cosh:

y = a*cosh((A^2-1)/(A^2+1) x) + b*(e^((A^2-1)/(A^2+1) x) - e^(-(A^2-1)/(A^2+1) x))/2

= (a+b/2)*cosh((A^2-1)/(A^2+1) x) + (b/2)*(e^((A^2-1)/(A^2+1) x) - e^(-(A^2-1)/(A^2+1) x))

We can simplify this further by combining the exponential terms:

y = A cosh((x+a)/A)

where A = sqrt((a+b/2)^2 + (b/2)^2) and a = (A^2-1)/(A^2+1) times the constant of integration, while b = -2a/(A^2+1) times the constant of integration.

Note that we have expressed a and b in terms of the constant of integration, which is determined by the initial conditions of the problem.