# `dy/dx = 1/ sqrt(80+8x-16x^2)` Solve the differential equation

## Expert Answers

For the given differential equation: `(dy)/(dx) = 1/sqrt(80+8x-16x^2),` we may write it in a form of `N(y) dy = M(x) dx` .

Cross-multiply the `(dx)` to the other side:

`(dy) = 1/sqrt(80+8x-16x^2) dx`

To solve for the general solution of the differential equation, we may apply direct integration on both...

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For the given differential equation: `(dy)/(dx) = 1/sqrt(80+8x-16x^2),` we may write it in a form of `N(y) dy = M(x) dx` .

Cross-multiply the `(dx)` to the other side:

`(dy) = 1/sqrt(80+8x-16x^2) dx`

To solve for the general solution of the differential equation, we may apply direct integration on both sides.

`int (dy) = int 1/sqrt(80+8x-16x^2) dx`

For the left side, it follow basic integral formula:

`int dy = y`

To evaluate the right side, we may apply completing the square on the trinomial: `80+8x-16x^2 = -(4x-1)^2+81`  or `81-(4x-1)^2`

Then, the integral on the right side becomes:

`int 1/sqrt(80+8x-16x^2) dx=int 1/sqrt(81-(4x-1)^2) dx`

The integral resembles the basic integration formula for inverse sine function:

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

We let `u = 4x-1` then `du = 4 dx` or `(du)/4= dx` .

Note that `81 = 9^2`

Then,

`int 1/sqrt(81-(4x-1)^2) dx =int 1/sqrt(9^2-u^2) *(du)/4`

`=(1/4)int1/sqrt(9^2-u^2)du`

`=(1/4)arcsin(u/9)+C`

Plug-in `u=4x-1` in `(1/4)arcsin(u/9` ), we get:

`int 1/sqrt(81-(4x-1)^2) dx= (1/4)arcsin((4x-1)/9) +C`

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

`y =1/4arcsin((4x-1)/9) +C`

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