The general solution of a differential equation in a form of `y’ =f(x,y)` can be 'evaluated using direct integration. The derivative of y denoted as ` y'` can be written as`(dy)/(dx) ` then `y'= f(x,y)` can be expressed as `(dy)/(dx)= f(x,y)` .

That is form of the given problem:`(dy)/(dx)=xsqrt(x-6)` .

We may apply the **variable separable differential** in which we follow `N(y) dy = M(x) dx` .

Cross-multiply `dx` to the right side: `dy=xsqrt(x-6)dx` .

Apply direct integration on both sides: `intdy=int xsqrt(x-6)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 = x-6` or `x = u+6` then `dx = du` .

`intxsqrt(x-6)dx=int (u+6)sqrt(u)du`

`=int (u+6)u^(1/2)du`

`=int(u^(3/2)+6u^(1/2))du`

Apply the basic integration property: `int (u+v) dx= int (u) dx + int (v) dx` .

`int u^(3/2)du+ int 6u^(1/2)du`

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

`int u^(3/2)du+ int 6u^(1/2)du=u^((3/2+1))/(3/2+1)+ 6u^((1/2+1))/(1/2+1)+C`

`=u^(5/2)/((5/2))+ 6u^(3/2)/((3/2))+C`

`=u^(5/2)*(2/5)+ 6u^(3/2)*(2/3)+C`

`=(2u^(5/2))/5+ 4u^(3/2)+C`

Plug-in `u = x-6` , we get:

`intxsqrt(x-6)dx=(2(x-6)^(5/2))/5+ 4(x-6)^(3/2)+C`

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

`y=(2(x-6)^(5/2))/5+ 4(x-6)^(3/2)+C`

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