For the given problem: `sqrt(x)+sqrt(y)y' =0,` we may rearrange this to

`sqrt(y)y' = -sqrt(x)`

Recall that `y'` is denoted as `(dy)/(dx)` then it becomes:

`sqrt(y)(dy)/(dx) = -sqrt(x)`

Apply the variable separable differential equation in a form of `f(y) dy = g(x) dx` .

`sqrt(y)(dy) = -sqrt(x)dx`

Apply direct integration using...

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For the given problem: `sqrt(x)+sqrt(y)y' =0,` we may rearrange this to

`sqrt(y)y' = -sqrt(x)`

Recall that `y'` is denoted as `(dy)/(dx)` then it becomes:

`sqrt(y)(dy)/(dx) = -sqrt(x)`

Apply the variable separable differential equation in a form of `f(y) dy = g(x) dx` .

`sqrt(y)(dy) = -sqrt(x)dx`

Apply direct integration using the Power Rule: `int u^n du = u^(n+1)/(n+1)` .

Note: `sqrt(x) = x^(1/2) and sqrt(y) = y^(1/2)` .

`int sqrt(y)(dy) = int -sqrt(x)dx`

`int y^(1/2) (dy) = int -x^(1/2)dx`

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

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

`y^(3/2)*2/3 = -x^(3/2)*2/3+C`

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

The general solution of the differential equation is `2/3y^(3/2)= -2/3x^(3/2)+C` .

Using the given initial condition `y(1)=9` , we plug-in `x=1` and `y=9` to solve for C:

`2/3(9)^(3/2)= -2/3*1^(3/2)+C`

`2/3*27=-2/3*1+C`

`18=-2/3+C`

`C = 18+2/3`

`C = 56/3`

So,

`2/3y^(3/2)= -2/3x^(3/2)+56/3`

`y^(3/2)=(3/2)(-2/3x^(3/2)+56/3)`

`y^(3/2)=-x^(3/2)+28`

`(y^(3/2))^(2/3)=(-x^(3/2)+28)^(2/3)

`y =(-x^(3/2)+28)^(2/3)`

or

y = `root(3)((-x^(3/2)+28)^2)`