Find the differential equation of the family of curves `y=A e^(2x)+B e^(-2x)` , where A and B are arbitrary constants.

Explanation

Let `y = 2/(sqrt(a^(2)-b^(2))) tan^(-1) {sqrt((a-b)/(a+b))tan""x/2}`
Differentiating it w.r.t x , we get
`(dy)/(dx) = 2/(sqrt(a^(2)-b^(2))). 1/(1+(a-b)/(a+b)tan ^(2)x//2).sqrt((a-b)/(a+b)).sec^(2)"" x/2 . 1/2`
` 2/(sqrt(a^(2)-b^(2))) .sqrt((a-b)/(a+b)). 1/2 . ((a+b)cos^(2)x//2 . sec^(2) x//2)/((a+b)cos^(2)x//2+ (a-b) sin^(2)x//2)`
` 1/(a(cos^(2)x//2 + sin^(2)x//2)+b(cos^(2)x//2 -sin^(2)x//2))`
`(dy)/(dx) = 1/(a + b cos x)`
` Rightarrow (d^(2)y)/(dx^(2)) = d/(dx)((dy)/(dx)) =-1/((a+b cos x)^(2)) (-b sin x) = (b sin x)/((a+ b cos x)^(2))`