# `sqrt(1-4x^2)y' = x` Find the general solution of the differential equation

*print*Print*list*Cite

To be able to evaluate the problem: `sqrt(1-4x^2)y'=x` , we express in a form of `y'=f(x)` .

To do this, we divide both sides by `sqrt(1-4x^2)` .

`y'=x/sqrt(1-4x^2)`

The general solution of a differential equation in a form of `y'=f(x)` can

be evaluated using direct integration. We can denote y' as `(dy)/(dx)` .

Then,

`y'=x/sqrt(1-4x^2)` becomes `(dy)/(dx)=x/sqrt(1-4x^2)`

This is the same as `(dy)=x/sqrt(1-4x^2) dx`

Apply direct integration on both sides:

For the left side, we have: `int (dy)=y`

For the right side, we apply u-substitution using `u =1-4x^2` then `du=-8x dx` or `(du)/(-8)=xdx` .

`int x/sqrt(1-4x^2) dx = int1/sqrt(u) *(du)/(-8)`

Applying basic integration property: `int c f(x) dx = c int f(x) dx` .

`int1/sqrt(u) *(du)/(-8) = -1/8int1/sqrt(u)du`

Applying Law of Exponents: `sqrt(x)= x^1/2` and `1/x^n = x^-n` :

`-1/8int1/sqrt(u)du=-1/8int1/u^(1/2)du`

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

Applying the Power Rule for integration: `int x^n= x^(n+1)/(n+1)+C` .

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

` =-1/8 u^(1/2)/(1/2)+C`

` =-1/8 u^(1/2)*(2/1)+C`

`= -2/8 u^(1/2)+C`

` = -1/4u^(1/2)+C or -1/4sqrt(u)+C`

Plug-in `u = 1-4x^2` in `-1/4u^(1/2)` , we get:

`int1/sqrt(u) *(du)/(-8)=-1/4sqrt(1-4x^2)+C`

Combining the results, we get the general solution for differential equation

`( sqrt(1-4x^2)y'=x)`

as:

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