Problem source

Note: You may assume that if the functions $y_1(x)$ and $y_2(x)$ both satisfy one of the differential equations in this question, then the curves $y = y_1(x)$ and $y = y_2(x)$ do not intersect.
\begin{questionparts}
\item Find the solution of the differential equation
$$\frac{dy}{dx} = y + x + 1$$
that has the form $y = mx + c$, where $m$ and $c$ are constants.
Let $y_3(x)$ be the solution of this differential equation with $y_3(0) = k$. Show that any stationary point on the curve $y = y_3(x)$ lies on the line $y = -x - 1$. Deduce that solution curves with $k < -2$ cannot have any stationary points.
Show further that any stationary point on the solution curve is a local minimum.
Use the substitution $Y = y + x$ to solve the differential equation, and sketch, on the same axes, the solutions with $k = 0$, $k = -2$ and $k = -3$.
\item Find the two solutions of the differential equation
$$\frac{dy}{dx} = x^2 + y^2 - 2xy - 4x + 4y + 3$$
that have the form $y = mx + c$.
Let $y_4(x)$ be the solution of this differential equation with $y_4(0) = -2$. (Do not attempt to find this solution.)
Show that any stationary point on the curve $y = y_4(x)$ lies on one of two lines that you should identify. What can be said about the gradient of the curve at points between these lines?
Sketch the curve $y = y_4(x)$. You should include on your sketch the two straight line solutions and the two lines of stationary points.
\end{questionparts}

Solution source

\begin{questionparts}
\item Looking for solution of the form $y = mx+c$ we find that $m = mx+c+x+1 \Rightarrow m = -1, c = -2$.

At stationary points $\frac{\d y}{\d x} = 0 \Rightarrow y = -x-1$.

If $y_3(0)= k < -2$ then the solution curve lies below $y = -x-2$ and therefore it cannot cross $y = -x -2$ to reach $y = -x-1$ for a stationary point.


Suppose $Y = y+x$ then $\frac{\d Y}{\d x} = \frac{\d y}{\d x} + 1=Y+2 \Rightarrow Y = Ae^x-2 \Rightarrow y= (k+2)e^x-x-2$


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){(#1)*(#1)+2*(#1)-1};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % 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
    }
    
    % 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);

        \foreach \a in {-15, ..., 15} {
        \draw[thin, red, dashed, domain=\xl:\xu, samples=100] 
            plot (\x, {(2+\a/3)*exp(\x)-\x-2});
        }
        
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {(2)*exp(\x)-\x-2});
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {(2-2)*exp(\x)-\x-2});
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {(2-3)*exp(\x)-\x-2});

        \draw[thick, green] (\xl, {-\xl-1})-- (\xu, {-\xu-1});
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

\item $\,$
\begin{align*}
&& m &= x^2 + (mx+c)^2 -2x(mx+c) - 4x+4(mx+c) + 3 \\
&&0&= (m^2-2m+1)x^2+(2mc-2c-4+4m)x + (c^2+4c+3-m)\\
\Rightarrow && m &= 1 \\
\Rightarrow && 0 &= c^2+4c+2 \\
\Rightarrow &&&= (c+2)^2-2 \\
\Rightarrow && c &= -2 \pm \sqrt{2}
\end{align*}

Therefore the lines are $y = x -2-\sqrt{2}$ and $y = x -2+\sqrt{2}$.

Any stationary point will satisfy $y' = 0$, ie $0 = x^2+y^2-2xy-4x+4y+3 = (x-y)^2-4(x-y)+3 = (x-y-3)(x-y-1)$ therefore they must lie on $y = x-1$ or $y = x-3$. Any point between these lines must have negative gradient (since one factor is positive and one factor is negative). 

\begin{center}
     \begin{tikzpicture}
    % 1. Mathematical Constants
    \def\sqrtTwo{1.4142}
    \def\lowerB{(-2 - \sqrtTwo)+0.0001} % Approx -3.4142
    \def\upperB{(-2 + \sqrtTwo)-0.0001} % Approx -0.5858
    
    % 2. Stable math function
    \def\soly(#1,#2){
        ((#1) - 2 - (2*tanh(\sqrtTwo*#1) - \sqrtTwo*(#2+2)) / 
        (\sqrtTwo - (#2+2)*tanh(\sqrtTwo*#1)))
    }
    
    \def\xl{-5} \def\xu{5}
    \def\yl{-10} \def\yu{10}
    
    \pgfmathsetmacro{\xscale}{10/(\xu-\xl)}
    \pgfmathsetmacro{\yscale}{10/(\yu-\yl)}
    
    \tikzset{
        axis/.style={very thick, ->},
        x=\xscale cm,
        y=\yscale cm
    }
    
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % 3. Family of curves between the equilibrium bounds
        % We iterate 12 times to keep the plot clean
        \foreach \a in {1, ..., 24} {
            % Calculate a y(0) value that is evenly spaced + a little random jitter
            % (rnd*0.2 - 0.1) adds a +/- 0.1 random shift
            \pgfmathsetmacro{\yzero}{\lowerB + (\a/25)*(\upperB-\lowerB)}
            
            \draw[thin, red, dashed, domain=\xl:\xu, samples=80] 
                plot (\x, {max(min(\soly(\x,\yzero), \yu+1), \yl-1)});
        }
        
        % The specific solution y(0) = -2 (exactly in the middle)
        \draw[thick, blue, domain=\xl:\xu, samples=100] 
            plot (\x, {\soly(\x,-2)});

        % The Equilibrium Bound Lines
        \draw[thick, blue!50!black, dashed] (\xl, {\xl+\upperB}) -- (\xu, {\xu+\upperB}) 
            node[pos=0.1, sloped, above] {\tiny $y = x-2+\sqrt{2}$};
        \draw[thick, blue!50!black, dashed] (\xl, {\xl+\lowerB}) -- (\xu, {\xu+\lowerB})
            node[pos=0.1, sloped, below] {\tiny $y = x-2-\sqrt{2}$};

        \draw[thick, green!50!black, dashed] (\xl, {\xl-3}) -- (\xu, {\xu-3}) 
            node[pos=0.1, sloped, above] {\tiny $y = x-3$};
        \draw[thick, green!50!black, dashed] (\xl, {\xl-1}) -- (\xu, {\xu-1})
            node[pos=0.1, sloped, below] {\tiny $y = x-1$};

    \end{scope}
    
    % Axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

\end{questionparts}