\(% 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}}\)

2018.2.8 Question 8

1. Since \(v = \sqrt {y}\), we have \(y = v^2\), and hence \[ \DiffFrac {y}{t} = 2v \DiffFrac {v}{t}, \] and hence the original equation reduces to \[ 2v \DiffFrac {v}{t} = \alpha v - \beta v^2, \] which gives \[ 2\DiffFrac {v}{t} = \alpha - \beta v. \]

   Rearranging gives us \[ \frac {\Diff v}{\alpha - \beta v} = \frac {\Diff t}{2}, \] and hence integrating both sides gives \[ -\frac {1}{\beta } \ln \abs *{\alpha - \beta v} = \frac {1}{2} t + C. \]

   Hence, \[ \ln \abs *{\alpha - \beta v} = -\frac {\beta t}{2} + C', \] and \[ \alpha - \beta v = A \exp \left (- \frac {\beta t}{2}\right ), \] and hence \[ v = \frac {1}{\beta } \left [\alpha + A \exp \left (- \frac {\beta }{2}\right )\right ], \] which means \[ y = v^2 = \frac {1}{\beta ^2} \left [\alpha + A \exp \left (- \frac {\beta t}{2}\right )\right ]^2. \]

   Since \(y = 0\) when \(t = 0\), we have \(A = - \alpha \), and hence \[ y_1(t) = \frac {\alpha ^2}{\beta ^2} \left [1 - \exp \left (- \frac {\beta t}{2}\right )\right ]^2. \]

2. Let \(v = \sqrt [3]{y}\) in this case, and hence \(y = v^3\), we have \[ \DiffFrac {y}{t} = 3v^2 \DiffFrac {v}{t}, \] and hence the original equation reduces to \[ 3v^2 \DiffFrac {v}{t} = \alpha v^2 - \beta v^3, \] and hence \[ 3 \DiffFrac {v}{t} = \alpha - \beta v. \]

   Similar to before, this solves to \[ v = \frac {1}{\beta } \left [\alpha + B \exp \left (- \frac {\beta }{3}\right )\right ], \] and hence \[ y = v^3 = \frac {1}{\beta ^3} \left [\alpha + B \exp \left (- \frac {\beta }{3}\right )\right ]^3. \]

   Since \(y = 0\) when \(t = 0\), we have \(B = -\alpha \), and hence \[ y_2(t) = \frac {\alpha ^3}{\beta ^3} \left [1 - \exp \left (- \frac {\beta t}{3}\right )\right ]^3. \]

3. Let \(\alpha = \beta = \gamma \). We have \[ y_1(t) = \left [1 - \exp \left (- \frac {\gamma t}{2}\right )\right ]^2, y_2(t) = \left [1 - \exp \left (- \frac {\gamma t}{3}\right )\right ]^3. \]

   For \(t > 0\), we have \[ 0 > -\frac {\gamma t}{3} > -\frac {\gamma t}{2} > -\infty , \] and since the exponential function is strictly increasing, we have \[ 1 > \exp \left (-\frac {\gamma t}{3}\right ) > \exp \left (-\frac {\gamma t}{2}\right ) > 0, \] and hence \[ 1 > 1 - \exp \left (-\frac {\gamma t}{2}\right ) > 1 - \exp \left (-\frac {\gamma t}{3}\right ) > 0. \]

   Hence, \[ y_1(t) = \left [1 - \exp \left (-\frac {\gamma t}{2}\right )\right ]^2 > \left [1 - \exp \left (-\frac {\gamma t}{3}\right )\right ]^2 > \left [1 - \exp \left (-\frac {\gamma t}{3}\right )\right ]^3 = y_2(t) \] which tells us that the graph of \(y_2\) should lie below the graph of \(y_1\).

   As \(t \to \infty \), \[ \exp \left (-\frac {\gamma t}{2}\right ), \exp \left (-\frac {\gamma t}{2}\right ) \to 0^{+}, \] and hence \[ y_1(t), y_2(t) \to 1^{-}. \]

   At \(t = 0\), \(y_1(t) = y_2(t) = 0\), and hence by the original differential equation \(y_1'(t) = y_2'(t) = 0\).

   Hence, the graph looks as follows.