For the given problem:`(dy)/(dx) =2xsqrt(4x^2+1)` is a first order ordinary differential equation in a form of `(dy)/(dx) = f(x,y)` .
To evaluate this, we rearrange it in a form of variable separable differential equation: `N(y) dy =M(x) dx` .
Cross-multiply `dx ` to the right side:`dy=2xsqrt(4x^2+1)dx` .
Apply direct integration on both sides: `intdy= int 2xsqrt(4x^2+1)dx` .
For the left side, we apply basic integration property: `int (dy)=y` .
For the right side, we may apply u-substitution by letting: `u = 4x^2+1` then `du=8x dx` or `(du)/8=x dx` .
The integral becomes:
`int 2xsqrt(4x^2+1)dx=int 2sqrt(u)*(du)/8`
`= int (sqrt(u)du)/4`
We may apply the basic integration property: `int c*f(x)dx= c int f(x) dx` .
`int (sqrt(u)du)/4= 1/4int sqrt(u)du`
Apply Law of Exponent: `sqrt(x)= x^(1/2)` and Power Rule for integration : int `x^n= x^(n+1)/(n+1)+C` .
`1/4int sqrt(u)du =(1/4) int u^(1/2)du`
`=(1/4)u^(1/2+1)/(1/2+1)+C`
`=(1/4)u^(3/2)/((3/2)) +C`
`=(1/4)u^(3/2)*(2/3) +C`
` =u^(3/2)/6+C`
Plug-in `u=4x^2+1` on `u^(3/2)/6+C` , we get:
`int 2xsqrt(4x^2+1)dx=(4x^2+1)^(3/2)/6+C`
Combining the results from both sides, we get the general solution of the differential equation as:
`y=(4x^2+1)^(3/2)/6+C`
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