Qualitative and Numerical Methods for Differential Equations

(a) $F(x,y)=x-y$, so the slopes are $0-0=0$, $1-0=1$, $0-1=-1$ and $2-1=1$.

(b) $x-y=k$ rearranges to $y=x-k$: a family of parallel lines of slope $1$, one per $k$. The nullcline $k=0$ is $y=x$, where every segment is horizontal.

(c) For $y=x-1$: left side $\frac{d}{dx}(x-1)=1$, right side $x-(x-1)=1$, equal for all $x$. For $y=x$: left side $1$, right side $x-x=0$, and $1\neq 0$.

$y_1=2+0.1(1)(2)=2.2$ at $x_1=1.1$, then

$y_2=2.2+0.1(1.1)(2.2)=2.2+0.242=2.442\quad\text{at } x_2=1.2 .$

Separating: $\frac{dy}{y}=x\,dx$, so $\ln y=\frac{x^2}{2}+C$ and $y=Ae^{x^2/2}$. From $y(1)=2$, $A=2e^{-1/2}$ and $y=2e^{(x^2-1)/2}$, giving $y(1.2)=2e^{0.22}\approx 2.4922$.

The error is $\approx 0.0502$. Differentiating the equation, $y''=y+xy'=y(1+x^2)>0$, so the curve is concave up and each tangent step lands below it.

(a) $f(y)=(y-2)(y+2)=0$ gives $y=\pm 2$. Signs: $f>0$ for $y<-2$, $f<0$ on $(-2,2)$, $f>0$ for $y>2$. Arrows point at $-2$ (stable) and away from $2$ (unstable). Check: $f'(y)=2y$, so $f'(-2)=-4<0$ and $f'(2)=4>0$.

(b) From $-5$: increasing, trapped below $-2$, so $y\to -2$. From $0$: decreasing toward the equilibrium beneath it, $y\to -2$. From $3$: increasing with nothing above it. Separating gives $y=\frac{10+2e^{4t}}{5-e^{4t}}$, which escapes at $t=\frac{\ln 5}{4}\approx 0.40$.

(a) $f(y)=0$ at $y=-2$ and at the double root $y=1$. Since $(y-1)^2\ge 0$, the sign of $f$ is the sign of $y+2$: $f<0$ for $y<-2$, and $f>0$ on both $(-2,1)$ and $(1,\infty)$. So arrows flee $-2$ (unstable) and point up on both sides of $1$ (semi-stable).

$f'(y)=2(y-1)(y+2)+(y-1)^2=3(y-1)(y+1)$, so $f'(-2)=9>0$ confirms unstable, while $f'(1)=0$ is the inconclusive repeated-root case.

(b) Follow the arrows: down from $-3$ without bound; up from $0$ to $1$; up from $1.5$ without bound.

(a) With $f(y)=y(4-y)$ and $f'(y)=4-2y$,

$y''=f'(y)f(y)=2(2-y)\,y\,(4-y).$

The solution starts at $y=1$ and uniqueness traps it in $(0,4)$, where $y>0$ and $4-y>0$, so the sign of $y''$ is the sign of $2-y$: concave up below the level $2$, concave down above it.

(b) This is logistic with $k=4$, $M=4$ and $A=\frac{M-y_0}{y_0}=3$, so $y=\frac{4}{1+3e^{-4t}}$. Setting $y=2$ gives $e^{-4t}=\frac13$, hence $t=\frac{\ln 3}{4}\approx 0.275$.

(a) $F(x,y)=y$ makes the update $y_{n+1}=y_n+hy_n=(1+h)y_n$, a geometric recursion, so $y_n=(1+h)^n$. Reaching $x=1$ takes $N=1/h$ steps, hence $y_N=(1+h)^{1/h}$.

(b) The exact value is $y(1)=e\approx 2.71828$. Then $2^1=2$, $1.5^2=2.25$, $1.25^4=2.44141$ and $1.125^8\approx 2.56578$; subtracting each from $e$ gives the errors, whose consecutive ratios are $1.53$, $1.69$, $1.82$.

(c) An error $\approx Ch^p$ would shrink by $2^p$ when $h$ halves. The ratios climb toward $2$, so $p=1$: first-order, $O(h)$. The values converge to $\lim_{h\to 0^+}(1+h)^{1/h}=e$.