Find (d²y)/(dx²) for: 1° arctan y = y-x; 2° ln y = y-x

Find $\dfrac{d^{2}y}{dx^{2}}$ for: 1° $\arctan y = y-x$; 2° $\ln y = y-x$; 3° $\sin\bigl(x+y\bigr) = y$.

Step 1: Part 1°: first derivative

\[ \frac{y'}{1+y^{2}} = y'-1 \;\Longrightarrow\; y'\left(\frac{1}{1+y^{2}}-1\right) = -1 \;\Longrightarrow\; y'\cdot\frac{-y^{2}}{1+y^{2}} = -1, \]

\[ y' = \frac{1+y^{2}}{y^{2}} = 1+\frac{1}{y^{2}}. \tag{1} \]

Step 2: Part 1°: second derivative

The right-hand form of (1) is the convenient one: differentiate $1+y^{-2}$ by the chain rule.

\[ y'' = \frac{d}{dy}\left(1+y^{-2}\right)\cdot y' = -\frac{2}{y^{3}}\cdot y' = -\frac{2}{y^{3}}\cdot\frac{1+y^{2}}{y^{2}}. \]

\[y'' = -\dfrac{2\bigl(1+y^{2}\bigr)}{y^{5}}\]

The sign is negative. It is easy to lose when the second derivative is extracted from the differentiated equation rather than from (1) directly; the check in Step 7 settles it numerically.

Step 3: Part 2°: first derivative

\[ \frac{y'}{y} = y'-1 \;\Longrightarrow\; y'\left(\frac1y-1\right) = -1 \;\Longrightarrow\; y' = \frac{y}{y-1} = 1+\frac{1}{y-1}. \tag{2} \]

Step 4: Part 2°: second derivative

\[ y'' = -\frac{1}{\bigl(y-1\bigr)^{2}}\cdot y' = -\frac{1}{\bigl(y-1\bigr)^{2}}\cdot\frac{y}{y-1}. \]

\[y'' = -\dfrac{y}{\bigl(y-1\bigr)^{3}}\]

Step 5: Part 3°: first derivative

Write $c = \cos\bigl(x+y\bigr)$ and $s = \sin\bigl(x+y\bigr)$ for brevity: \[c\bigl(1+y'\bigr) = y' \;\Longrightarrow\; c = y'\bigl(1-c\bigr) \;\Longrightarrow\; y' = \frac{c}{1-c}.\]

A form that will be needed in a moment: \[1+y' = \frac{\bigl(1-c\bigr)+c}{1-c} = \frac{1}{1-c}. \tag{3}\]

Step 6: Part 3°: second derivative

Differentiate $c\bigl(1+y'\bigr) = y'$, noting that $c' = -s\bigl(1+y'\bigr)$: \[-s\bigl(1+y'\bigr)^{2}+cy'' = y'' \;\Longrightarrow\; y''\bigl(c-1\bigr) = s\bigl(1+y'\bigr)^{2}.\]

Using (3) and $c-1 = -\bigl(1-c\bigr)$: $y'' = \frac{s}{\bigl(c-1\bigr)\bigl(1-c\bigr)^{2}} = -\frac{s}{\bigl(1-c\bigr)^{3}}$.

\[y'' = -\dfrac{\sin\bigl(x+y\bigr)}{\Bigl[1-\cos\bigl(x+y\bigr)\Bigr]^{3}}\]

Step 7: Verification

\[1^{\circ}\ \text{at } y = 1:\ x = 1-\frac{\pi}{4} \approx 0.2146019,\ y' = 2,\ y'' = -4\]

\[\text{Numerically, solving } \arctan y = y-x \text{ at } x_{0}\pm10^{-3}:\ \frac{y_{+}-2y_{0}+y_{-}}{10^{-6}} \approx -4.00003 \;\;\checkmark\]

\[2^{\circ}\ \text{at } y = 2:\ x = 2-\ln2 \approx 1.3069,\ y' = 2,\ y'' = -2 \;\;\checkmark \qquad 3^{\circ}:\ \text{symbolic residual } 0 \;\;\checkmark\]