\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We have \begin {align*} I - I_0 & = \int _{0}^{1} \left [(y')^2 - y^2 - (y' + y \tan x)^2\right ] \Diff x \\ & = - \int _{0}^{1} \left [y^2 + y^2 \tan ^2 x + 2y y' \tan x\right ] \Diff x \\ & = - \int _{0}^{1} \left [y^2 (1 + \tan ^2 x) + 2y y' \tan x \right ] \Diff x \\ & = - \int _{0}^{1} \left (y^2 \cdot \sec ^2 x + 2y \cdot y' \cdot \tan x\right ) \Diff x. \end {align*}
But notice that \[ \DiffOp {x} y^2 \tan x = y^2 \cdot \sec ^2 x + 2y \cdot y' \cdot \tan x, \] and hence \begin {align*} I - I_0 & = - \int _{0}^{1} \left (y^2 \cdot \sec ^2 x + 2y \cdot y' \cdot \tan x\right ) \Diff x \\ & = - \left [y^2 \tan x\right ]_{0}^{1} \\ & = - (y(1)^2 \tan 1 - 0^2 \tan 0) \\ & = - (0^2 \tan 1 - 0) \\ & = 0, \end {align*}
as desired.
This gives \(I = I_1\). Also, notice that the integrand of \(I_1\) is \((y' + y \tan x)^2\) is always non-negative, which means \(I_1 \geq 0\), taking \(0\) only when \(y' + y \tan x = 0\) for all \(x \in (0, 1)\).
\begin {align*} y' + y \tan x & = 0 \\ \DiffFrac {y}{x} & = -y \tan x \\ \frac {\Diff y}{y} & = - \tan x \Diff x \\ \ln \abs *{y} & = - \ln \abs *{\sec x} + C \\ y & = A\cos x. \\ \end {align*}
When \(x = 1\), \(y = 0\), hence \(A = 0\) since \(\cos 1 \neq 0\). This means \(I_1 = 0\) if and only if \(y = 0\) for all \(x \in [0, 1]\).
Since \(I = I_1\), we know that \(I \geq 0\), with the equal sign holding if and only if \(y = 0\) for all \(x \in [0, 1]\).
Let \[ J_0 = \int _{0}^{1} (y' + ay\tan bx)^2 \Diff x, \] and we have \begin {align*} J - J_0 & = \int _{0}^{1} \left [((y')^2 - a^2 y^2) - (y' + ay\tan bx)^2\right ] \Diff x \\ & = -\int _{0}^{1} \left [a^2 y^2 + a^2 y^2 \tan ^2 bx + 2y' \cdot y \cdot a \cdot \tan bx \right ] \Diff x \\ & = -\int _{0}^{1} \left [a^2 y^2 \sec ^2 bx + 2y' \cdot y \cdot a \cdot \tan bx \right ] \Diff x \\ & = -a \int _{0}^{1} \left [a y^2 \sec ^2 bx + 2y' \cdot y \cdot \tan bx \right ] \Diff x. \end {align*}
Notice that if we let \(b = a\), we have \[ \DiffFrac {y^2 \tan bx}{x} = 2y y' \tan bx + b y^2 \sec ^2 bx = 2y y' \tan bx + a y^2 \sec ^2 bx. \]
This means \begin {align*} J - J_0 & = -a \int _{0}^{1} \left [a y^2 \sec ^2 bx + 2y' \cdot y \cdot \tan bx \right ] \Diff x \\ & = -a \left [y^2 \tan ax\right ]_{0}^{1} \\ & = -a (y(1)^2 \tan a - 0^2 \tan 0) \\ & = 0. \end {align*}
This means \(J = J_0\).
Since the integrand of \(J_0\) is a square, we know \(J_0 \geq 0\) and hence \(J \geq 0\).
This is only valid when \(ax < \frac {\pi }{2}\) for \(x \in [0, 1]\) (since otherwise this range will cross an undefined point), which means \(a < \frac {\pi }{2}\).
When \(a = \frac {\pi }{2}\), consider \(y = \cos ax\). Notice that \(y' = - a \sin ax\), and therefore \begin {align*} J & = \int _{0}^{1} ((-a \sin ax)^2 - a^2 \cos ^2 ax) \Diff x \\ & = -a^2 \int _{0}^{1} (\cos ^2 ax - \sin ^2 ax) \Diff x \\ & = -a^2 \int _{0}^{1} \cos (2ax) \Diff x \\ & = -a^2 \left [\frac {\sin 2ax}{2a}\right ]_{0}^{1} \\ & = -\frac {a}{2} \left [\sin \pi x\right ]_{0}^{1} \\ & = -\frac {a}{2} (0 - 0) \\ & = 0, \end {align*}
but \(y\) is not uniformly zero.