The solution of differential equation. day/dx = 1/√1+x^2 is y =?

The solution of differential equation. day/dx = 1/√1+x^2 is y =?

Explanation

The given differential equation is:

dy/dx = 1/√(1+x^2)

To solve this equation, we can use the substitution u = 1+x^2, which implies du/dx = 2x. Then, we can rewrite the equation as:

dy/dx = 1/√u

dy = 1/√u du/2x

2y = ∫1/√u du

Now, we can recognize the integral as the inverse hyperbolic sine (arsinh or sinh^-1) function:

2y = sinh^-1(u) + C

2y = sinh^-1(1+x^2) + C

Dividing both sides by 2, we get:

y = (1/2)sinh^-1(1+x^2) + C/2

Since C/2 is also a constant, we can write the solution as:

y = (1/2)sinh^-1(1+x^2) + c

or simply: y = sinh^-1(x) + c