Solve the separable differential equation...?

2x - 4y√(1+x^2) dy/dx = 0
- 4y √(1+x^2) dy/dx = - 2x
....................- 2xdx
...- 4∫ydy = -------------
...............√(1+x^2)

let u = 1 + x^2
du = 2xdx

- 4∫ydy = - ∫ u^(-1/2)du
- 2y^2 = - 2u^(1/2) + C

substitute back for u = 1 + x^2

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

when y(0) = 2

- 2(2)^2 = - 2[1 + (0)^2]^(1/2) + C
- 8 = - 2(1) + C
- 8 = - 2 + C
- 8 + 2 = + C
C = - 6


- 2y^2 = - 2(1 + x^2)^(1/2) - 6

y^2 =√(1 + x^2) + 3
y = √[√(x^2 + 1) + 3] answer//

Du/dt = e^(3u) x e^(2t) du/e^3u = e^2t dt ie e^-3u du = e ^2t dt by using integrating, -1/three e^-3u = 1/2e^2t + C when t= 0 u =17 -1/3e^-fifty one = half + C - 1/three e^-3u = half e^2t - 1/3e^-51-1/2 e^-3u = e^-fifty one +three/2 -three/2e^2t

2x - 4y√(1+x^2) dy/dx = 0

2x = 4y√(1+x^2) dy/dx

Separate the variables
2x dx / √(1+x^2) = 4 y dy
4 y dy = 2x dx / √(1+x^2)
2 y dy = x dx / √(1+x^2)

2 ∫ y dy = ∫ x dx / √(1+x^2) ----- (1)

∫ x dx / √(1+x^2)
Let u= 1+x^2
du = 2x dx
x dx = (1/2) du

∫ x dx / √(1+x^2) = (1/2) ∫ du/ sqrt(u) = (1/2) ∫ u^(-1/2) du = (1/2) u^(-1/2+1)/(-1/2+1)
= (1/2) (2) u^(1/2) = u^(1/2) = (1+x^2)^(1/2) = √(1+x^2)

(1) becomes:
2 ∫ y dy = = √(1+x^2)
y^2 = √(1+x^2) + C
y(0)=2
4 = 1+C
C=3

y^2 = √(1+x^2) + 3
y = sqrt ( √(1+x^2) + 3)

2x - 4y sqrt(1 + x^2) (dy/dx) = 0
-4y sqrt(1 + x^2) (dy/dx) = -2x
-4y sqrt(1 + x^2) dy = -2x dx
-4y dy = -2x/sqrt(1 + x^2) dx
2y dy = x/sqrt(1 + x^2) dx

Now take the integral of both sides.

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

Let u = 1 + x^2.
du = 2x dx
(1/2) du = x dx

y^2 = Integral ( 1/sqrt(u) (1/2) du )
y^2 = (1/2) Integral ( 1/u^(1/2) du )
y^2 = (1/2) (2u^(1/2)) + C
y^2 = u^(1/2) + C
y^2 = (1 + x^2)^(1/2) + C

In order for the function to be subject to the condition y(0) = 2, it means (0, 2) is a solution. Plug in x = 0 and y = 2 to solve for C.

2^2 = (1 + 0^2)^(1/2) + C
4 = 1 + C
3 = C

y^2 = (1 + x^2)^(1/2) + 3

Separating variables gives

2y dy = [x/√(1+x^2)] dx
Integrating gives

y^2 = √(1+x^2) + C
y=2 when x=0 gives C=3
so y^2 = √(1+x^2) + 3 giving
y = ± √[√(1+x^2)+3]