Problem source

It is given that $y$ satisfies
 $$
{{\d y} \over { \d t}} + 
k\left({{t^2-3t+2} \over {t+1}}\right)y = 0\;,
$$
where $k$ is a constant,  and $y=A $ when $t=0\,$, where $A$ is a positive constant.
Find $y$ in terms of $t\,$, $k$ and $A\,$.
Show that $y$ has two stationary values whose 
ratio is  $(3/2)^{6k}\e^{-5{k}/2}.$
Describe the behaviour of $y$ as $t \to +\infty$ for the case where $k> 0$ and for the case where $k<0\,.$
In separate diagrams, sketch the graph of $y$  for $t>0$ for each of these cases.

Solution source

\begin{align*}
&& \frac{\d y}{\d t} &= - k \left (\frac{t^2-3t+2}{t+1} \right) y \\
\Rightarrow && \int \frac1y \d y &= -k\int \left (t-4 + \frac{6}{t+1}\right) \d t \\
\Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + C \\
(t,y) = (0,A): && \ln A &=C \\
\Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + \ln A \\
&& \ln \left ( \frac{y}{A}(t+1)^{6k} \right) &= -k \l \frac12 t^2 - 4t \r \\
\Rightarrow && y &= A\frac{\exp \l -k(\frac12 t^2-4t)\r}{(t+1)^{6k}}
\end{align*}

$y$ wil have stationary values when $\frac{\d y}{\d t} = 0$, ie
\begin{align*}
k \left (\frac{t^2-3t+2}{t+1} \right) y &= 0 \\
k \left ( \frac{(t-2)(t-1)}{t+1} \right) y &= 0
\end{align*}

ie when $y = 0, t = 1, t =2$. Clearly $y = 0$ is not a solution, so $y$ has the values:

\begin{align*}
t = 1: && y &= A\frac{\exp \l -k(\frac12 -4)\r}{(2)^{6k}} \\
&&&= A \frac{e^{7/2 k}}{2^{6k}} \\
t = 2: &&  y &= A\frac{\exp \l -k(2 -8)\r}{(3)^{6k}} \\
&&&= A \frac{e^{6 k}}{3^{6k}} \\
\text{ratio}: && \frac{e^{7/2k}}{2^{6k}} \cdot \frac{3^{6k}}{e^{6k}} &= (3/2)^{6k} e^{-5k/2}
\end{align*} 

If $k > 0$ as $t \to \infty$ $y \to 0$ since the $e^{-kt^2/2}$ term dominates everything.

If $k < 0$ as $t \to \infty$ $y \to \infty$ as since the $e^{-kt^2}$ term also dominates but now it growing to infinity faster than everything else.


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){exp(0.5*(#1)^2-4*(#1)-6*ln(1+(#1)))};
    \def\functiong(#1){(0.0125051796)*(#1)^0+(-0.1647073118)*(#1)^1+(1.0119000852)*(#1)^2+(-3.1771944633)*(#1)^3+(2.6800185961)*(#1)^4+(9.7349942102)*(#1)^5+(-19.7050808814)*(#1)^6+(-8.8929259572)*(#1)^7+(38.1744856073)*(#1)^8+(-1.117138306)*(#1)^9+(-31.5879434899)*(#1)^10+(2.6621658188)*(#1)^11+(8.8345410319)*(#1)^12+(4.0085358261)*(#1)^13+(1.0787869889)*(#1)^14+(-3.5532424704)*(#1)^15};
    \def\xl{-1};
    \def\xu{11};
    \def\yl{-0.5};
    \def\yu{1};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    % 3 - 9/x + 27/x^2 - 147/(2 x^3) + 198/x^4 - 4275/(8 x^5) + 1443/x^6 - 62343/(16 x^7) + 10521/x^8 - 3636267/(128 x^9) + O((1/x)^10)
% (Laurent series)
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=0:11, samples=100] 
            plot ({\x}, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=-1:1, samples=100] 
            plot ({5*(\x+1)}, {\functiong(\x)}); 

        % \draw[thick, blue, smooth, domain=0.01:0.1, samples=100]
            % plot(1/\x, {3 - 9*\x + 27*\x^2 - 147/2 *\x^3 + 198*\x^4-4275/8 * \x^5 });
        % \draw[thick, red, dashed, smooth, domain=0.1:25, samples=100] 
        %     plot (\x, {3-9/\x});  
        % \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        % \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        \draw[red] (25,3.25) node [below left] {$y = 3-\frac9x$};
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$z$};
    
    \end{tikzpicture}
\end{center}