`dy/dx = 1/((x-1)sqrt(-4x^2+8x-1))` Solve the differential equation

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 The given problem` (dy)/(dx) =1/((x-1)sqrt(-4x^2+8x+1)) ` is in form of a first order ordinary differential equation. To evaluate this, we may follow the variable separable differential equation: `N(y) dy= M(x)dx` .

`dy=1/((x-1)sqrt(-4x^2+8x+1)) dx`

Apply direct integration on both sides:

`int dy=int 1/((x-1)sqrt(-4x^2+8x+1)) dx`

For the left side, we apply basic integration property: `int (dy)=y.`

For the right side, we apply several substitutions to simplify it.

 Let `u =(x-1)` then `x=u+1` and `du=dx` . The integral becomes:

`int 1/((u)sqrt(-4x^2+8x+1)) dx =int 1/(usqrt(-4(u+1)^2+8(u+1)+1)) du`

`=int 1/(usqrt(-4(u^2+2u+1)+8u+8+1)) du`

`=int 1/(usqrt(-4u^2-8u-4+8u+8+1)) du`

`=int 1/(usqrt(-4u^2+5)) du`

Let `v = u^2` then `dv = 2u du` or `(dv)/(2u)=du` . The integral becomes:

`int 1/(usqrt(-4u^2+5)) du=int 1/(usqrt(-4v+5)) *(dv)/(2u)`

`=int (dv)/(2u^2sqrt(-4v+5))`

`=int (dv)/(2vsqrt(-4v+5))`

Apply the basic integration property: `int c*f(x)dx= c int f(x) dx` .

`int (dv)/(2vsqrt(-4v+5)) =(1/2)int (dv)/(vsqrt(-4v+5))`

Let `w= sqrt(-4v+5)` then `v= (5-w^2)/4` and `dw=-2/sqrt(-4v+5)dv` or

`(dw)/(-2)=1/sqrt(-4v+5)dv`

The integral becomes:

`(1/2)int (dv)/(vsqrt(-4v+5)) =(1/2)int 1/v*(dv)/sqrt(-4v+5)`

`=(1/2)int 1/((5-w^2)/4)*(dw)/(-2)`

`=(1/2)int 1*4/(5-w^2)*(dw)/(-2)`

`=(1/2)int -2/(5-w^2)dw`

`=(1/2)*-2 int 1/(5-w^2)dw`

`=(-1) int 1/(5-w^2)dw`

Apply basic integration formula for inverse hyperbolic tangent function:

`int (du)/(a^2-u^2)=(1/a)arctanh(u/a)+C`

Then, with corresponding values as: `a^2=5` and  `u^2=u^2` , we get: `a=sqrt(5)` and `u=w`  

`(-1) int 1/(5-w^2)dw = -1/sqrt(5) arctanh(w/sqrt(5))+C`

Recall `w=sqrt(-4v+5)`  and `v=u^2` then `w =sqrt(-4u^2+5).`

Plug-in `u=(x-1)` on `w =sqrt(-4u^2+5)` , we get:

`w =sqrt(-4(x-1)^2+5)`

`w=sqrt(-4(x^2-2x+1)+5)`

`w=sqrt(-4x^2+8x-4+5)`

`w=sqrt(-4x^2+8x+1)`

 

Plug-in `w=sqrt(-4x^2+8x+1)` on `-1/sqrt(5) arctanh(w/sqrt(5))+C` , we get:

`int 1/((x-1)sqrt(-4x^2+8x+1)) dx=1/sqrt(5)arctanh(sqrt(-4x^2+8x+1)/sqrt(5))+C`

`=-1/sqrt(5) arctanh(sqrt(-4x^2+8x+1)/5)+C`

Combining the results from both sides, we get the general solution of the differential equation as:

`y=-1/sqrt(5) arctanh(sqrt(-4x^2+8x+1)/5)+C`

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