It is given that $\frac{\text{d}y}{\text{d}x} = \frac{x+5}{y^2}$ and when $x=0$, $y=5$. Find the Maclaurin's series expansion for $y$, up to and including the term in $x^3$, leaving the coefficients in exact form.

1. Evaluate the first derivative at $x=0$.

Given the differential equation:
$\frac{\text{d}y}{\text{d}x} = \frac{x+5}{y^2}$
Substitute $x=0$ and $y=5$:
$\frac{\text{d}y}{\text{d}x} \bigg|_{x=0} = \frac{0+5}{5^2} = \frac{1}{5}$

2. Rearrange and differentiate the equation implicitly to find an expression for the second derivative.

Rearranging the equation to avoid the quotient rule:
$y^2 \frac{\text{d}y}{\text{d}x} = x + 5$
Differentiating implicitly with respect to $x$ using the product and chain rules:
$2y \left( \frac{\text{d}y}{\text{d}x} \right)^2 + y^2 \frac{\text{d}^2y}{\text{d}x^2} = 1$

3. Evaluate the second derivative at $x=0$.

Substitute $x=0$, $y=5$, and $\frac{\text{d}y}{\text{d}x} = \frac{1}{5}$ into the second derivative equation:
$2(5)\left(\frac{1}{5}\right)^2 + (5)^2 \frac{\text{d}^2y}{\text{d}x^2} = 1$
$10\left(\frac{1}{25}\right) + 25 \frac{\text{d}^2y}{\text{d}x^2} = 1$
$\frac{2}{5} + 25 \frac{\text{d}^2y}{\text{d}x^2} = 1$
$25 \frac{\text{d}^2y}{\text{d}x^2} = \frac{3}{5}$
$\frac{\text{d}^2y}{\text{d}x^2} \bigg|_{x=0} = \frac{3}{125}$

4. Differentiate implicitly again to find an expression for the third derivative.

Differentiate $2y \left( \frac{\text{d}y}{\text{d}x} \right)^2 + y^2 \frac{\text{d}^2y}{\text{d}x^2} = 1$ with respect to $x$:
$2 \frac{\text{d}y}{\text{d}x} \left( \frac{\text{d}y}{\text{d}x} \right)^2 + 2y \left( 2 \frac{\text{d}y}{\text{d}x} \frac{\text{d}^2y}{\text{d}x^2} \right) + 2y \frac{\text{d}y}{\text{d}x} \frac{\text{d}^2y}{\text{d}x^2} + y^2 \frac{\text{d}^3y}{\text{d}x^3} = 0$
Simplifying the terms:
$2 \left( \frac{\text{d}y}{\text{d}x} \right)^3 + 6y \frac{\text{d}y}{\text{d}x} \frac{\text{d}^2y}{\text{d}x^2} + y^2 \frac{\text{d}^3y}{\text{d}x^3} = 0$

5. Evaluate the third derivative at $x=0$.

Substitute $y=5$, $\frac{\text{d}y}{\text{d}x} = \frac{1}{5}$, and $\frac{\text{d}^2y}{\text{d}x^2} = \frac{3}{125}$:
$2 \left( \frac{1}{5} \right)^3 + 6(5)\left( \frac{1}{5} \right)\left( \frac{3}{125} \right) + (5)^2 \frac{\text{d}^3y}{\text{d}x^3} = 0$
$2 \left( \frac{1}{125} \right) + 6 \left( \frac{3}{125} \right) + 25 \frac{\text{d}^3y}{\text{d}x^3} = 0$
$\frac{2}{125} + \frac{18}{125} + 25 \frac{\text{d}^3y}{\text{d}x^3} = 0$
$\frac{20}{125} + 25 \frac{\text{d}^3y}{\text{d}x^3} = 0$
$25 \frac{\text{d}^3y}{\text{d}x^3} = -\frac{4}{25}$
$\frac{\text{d}^3y}{\text{d}x^3} \bigg|_{x=0} = -\frac{4}{625}$

6. Construct the Maclaurin's series expansion using the evaluated derivatives.

The Maclaurin's series formula is:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
Substituting the values obtained:
$y = 5 + \left(\frac{1}{5}\right)x + \frac{\left(\frac{3}{125}\right)}{2}x^2 + \frac{\left(-\frac{4}{625}\right)}{6}x^3 + \dots$
$y = 5 + \frac{1}{5}x + \frac{3}{250}x^2 - \frac{4}{3750}x^3 + \dots$
$y = 5 + \frac{1}{5}x + \frac{3}{250}x^2 - \frac{2}{1875}x^3$