Given that $y = \text{e}^{\sqrt{1+2x}}$, show that
$(1+2x)\left(\frac{\text{d}y}{\text{d}x}\right)^2 = y^2 \quad \text{and} \quad (1+2x)\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} = y.$ Hence, or otherwise, obtain the series expansion for $y$ in terms of $x$ up to and including the term in $x^3$. Verify that the same series expansion for $y$ in part (b) is obtained if the standard series expansions for $\text{e}^x$ and $(1+x)^n$ are used.

1. Find the first derivative and square both sides to obtain the first result.

$y = \text{e}^{(1+2x)^{1/2}}$
$\frac{\text{d}y}{\text{d}x} = \text{e}^{(1+2x)^{1/2}} \cdot \frac{1}{2}(1+2x)^{-1/2}(2)$
$\frac{\text{d}y}{\text{d}x} = \frac{y}{\sqrt{1+2x}}$
$\sqrt{1+2x}\frac{\text{d}y}{\text{d}x} = y$
Squaring both sides:
$(1+2x)\left(\frac{\text{d}y}{\text{d}x}\right)^2 = y^2 \quad \text{(shown)}$

2. Differentiate the first result implicitly with respect to $x$ using the product and chain rules.

$\frac{\text{d}}{\text{d}x} \left[ (1+2x)\left(\frac{\text{d}y}{\text{d}x}\right)^2 \right] = \frac{\text{d}}{\text{d}x} \left( y^2 \right)$
$2\left(\frac{\text{d}y}{\text{d}x}\right)^2 + (1+2x)\left( 2\frac{\text{d}y}{\text{d}x}\frac{\text{d}^2y}{\text{d}x^2} \right) = 2y\frac{\text{d}y}{\text{d}x}$

3. Factor out common terms and simplify to obtain the second given result.

$2\frac{\text{d}y}{\text{d}x} \left( \frac{\text{d}y}{\text{d}x} + (1+2x)\frac{\text{d}^2y}{\text{d}x^2} \right) = 2y\frac{\text{d}y}{\text{d}x}$
Since $y = \text{e}^{\sqrt{1+2x}}$ is strictly positive, $\frac{\text{d}y}{\text{d}x} \neq 0$. Dividing both sides by $2\frac{\text{d}y}{\text{d}x}$ gives:
$\frac{\text{d}y}{\text{d}x} + (1+2x)\frac{\text{d}^2y}{\text{d}x^2} = y$
$(1+2x)\frac{\text{d}^2y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} = y \quad \text{(shown)}$

1. Evaluate $y$ and $\frac{\text{d}y}{\text{d}x}$ at $x=0$.

Given $y = \text{e}^{\sqrt{1+2x}}$.
When $x=0$, $y = \text{e}^{\sqrt{1}} = \text{e}$.

Differentiating directly to find $\frac{\text{d}y}{\text{d}x}$ at $x=0$:
$\frac{\text{d}y}{\text{d}x} = \text{e}^{\sqrt{1+2x}} \cdot \frac{1}{2}(1+2x)^{-\frac{1}{2}}(2) = \frac{\text{e}^{\sqrt{1+2x}}}{\sqrt{1+2x}}$
When $x=0$, $\frac{\text{d}y}{\text{d}x} = \frac{\text{e}}{1} = \text{e}$.

2. Use the differential equation from part (a) to evaluate $\frac{\text{d}^2 y}{\text{d}x^2}$ at $x=0$.

From part (a), we are given:
$(1+2x)\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} = y$
Substitute $x=0$, $y=\text{e}$, and $\frac{\text{d}y}{\text{d}x}=\text{e}$:
$(1+0)\frac{\text{d}^2 y}{\text{d}x^2} + \text{e} = \text{e}$
$\frac{\text{d}^2 y}{\text{d}x^2} = 0$

3. Differentiate the given differential equation with respect to $x$ to find $\frac{\text{d}^3 y}{\text{d}x^3}$ at $x=0$.

Differentiate the equation $(1+2x)\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} = y$ implicitly with respect to $x$ using the product rule:
$(1+2x)\frac{\text{d}^3 y}{\text{d}x^3} + 2\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}^2 y}{\text{d}x^2} = \frac{\text{d}y}{\text{d}x}$
$(1+2x)\frac{\text{d}^3 y}{\text{d}x^3} + 3\frac{\text{d}^2 y}{\text{d}x^2} = \frac{\text{d}y}{\text{d}x}$
Substitute $x=0$, $\frac{\text{d}^2 y}{\text{d}x^2}=0$, and $\frac{\text{d}y}{\text{d}x}=\text{e}$:
$(1+0)\frac{\text{d}^3 y}{\text{d}x^3} + 3(0) = \text{e}$
$\frac{\text{d}^3 y}{\text{d}x^3} = \text{e}$

4. Substitute the evaluated derivatives into the Maclaurin series formula.

Using the Maclaurin series expansion formula:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
Substitute the known values:
$y = \text{e} + (\text{e})x + \frac{0}{2!}x^2 + \frac{\text{e}}{3!}x^3 + \dots$
$y \approx \text{e} + \text{e}x + \frac{\text{e}}{6}x^3$

1. Expand the exponent $\sqrt{1+2x}$ using the standard binomial series up to the term in $x^3$.

Using the binomial expansion for $(1+u)^n$:
$\sqrt{1+2x} = (1+2x)^{\frac{1}{2}}$
$= 1 + \left(\frac{1}{2}\right)(2x) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(2x)^2 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(2x)^3 + \dots$
$\approx 1 + x - \frac{1}{8}(4x^2) + \frac{1}{16}(8x^3)$
$= 1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3$

2. Substitute the expanded exponent into $y$ and factor out the constant term to prepare for the $\text{e}^X$ series.

$y = \text{e}^{\sqrt{1+2x}} \approx \text{e}^{1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3}$
To use the standard Maclaurin series for $\text{e}^X$ (which requires $X \to 0$ as $x \to 0$), we factor out $\text{e}^1$:
$y = \text{e}^1 \cdot \text{e}^{x - \frac{1}{2}x^2 + \frac{1}{2}x^3}$

3. Apply the standard series expansion for $\text{e}^X$.

Using the standard series $\text{e}^X = 1 + X + \frac{X^2}{2!} + \frac{X^3}{3!} + \dots$, let $X = x - \frac{1}{2}x^2 + \frac{1}{2}x^3$:
$y \approx \text{e} \left[ 1 + \left(x - \frac{1}{2}x^2 + \frac{1}{2}x^3\right) + \frac{1}{2}\left(x - \frac{1}{2}x^2\right)^2 + \frac{1}{6}\left(x\right)^3 \right]$
(Note: Higher powers in the expansion of $X^2$ and $X^3$ are omitted as they result in terms of $x^4$ and above.)

4. Expand, collect like terms up to $x^3$, and verify the result.

$\left(x - \frac{1}{2}x^2\right)^2 \approx x^2 - x^3$
Substitute back into the expression for $y$:
$y \approx \text{e} \left[ 1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3 + \frac{1}{2}(x^2 - x^3) + \frac{1}{6}x^3 \right]$
$y \approx \text{e} \left[ 1 + x - \frac{1}{2}x^2 + \frac{1}{2}x^3 + \frac{1}{2}x^2 - \frac{1}{2}x^3 + \frac{1}{6}x^3 \right]$
$y \approx \text{e} \left( 1 + x + \frac{1}{6}x^3 \right) = \text{e} + \text{e}x + \frac{\text{e}}{6}x^3$
This matches the series expansion obtained in part (b), thus verifying the result.