Differential Equations

Part (i). Set $v=\sqrt{y}$, so $y=v^2$ and $\frac{dy}{dt}=2v\frac{dv}{dt}$. Substituting into the equation gives $$2v\frac{dv}{dt}=\alpha v-\beta v^2,$$ and since $y⩾0$ we may divide through by $v>0$ (we handle $v=0$ separately) to obtain $$2\frac{dv}{dt}=\alpha -\beta v.$$ Separating variables: $$\frac{2dv}{\alpha -\beta v}=dt.$$ Integrating both sides: $$-\frac{2}{\beta }\ln |\alpha -\beta v|=t+C.$$ Applying the initial condition $y_1(0)=0$ means $v(0)=0$, giving $C=-\frac{2}{\beta }\ln \alpha$. Hence $$\ln \left(\frac{\alpha -\beta v}{\alpha }\right)=-\frac{\beta t}{2},\text{so}\alpha -\beta v=\alpha e^{-\beta t/2}.$$ Solving for $v$: $v=\frac{\alpha }{\beta }\left(1-e^{-\beta t/2}\right)$, and squaring back: $$\boxed{y_1(t)=\frac{{\alpha }^2}{{\beta }^2}{\left(1-e^{-\beta t/2}\right)}^2.}$$ This is non-constant and satisfies $y_1(0)=0$. As $t\to \infty$, $y_1\to {\alpha }^2/{\beta }^2$, a horizontal asymptote. Part (ii). The structure is identical with exponent $\frac{2}{3}$ instead of $\frac{1}{2}$. Try $v=y^{1/3}$, so $y=v^3$ and $\frac{dy}{dt}=3v^2\frac{dv}{dt}$. Substituting: $$3v^2\frac{dv}{dt}=\alpha v^2-\beta v^3=v^2(\alpha -\beta v),$$ and dividing by $v^2>0$: $$3\frac{dv}{dt}=\alpha -\beta v.$$ This is the same structure as part (i) with the factor $3$ replacing $2$. Separating and integrating with $v(0)=0$ gives $v=\frac{\alpha }{\beta }\left(1-e^{-\beta t/3}\right)$, so $$\boxed{y_2(t)=\frac{{\alpha }^3}{{\beta }^3}{\left(1-e^{-\beta t/3}\right)}^3.}$$ Part (iii). When $\alpha =\beta$, the two solutions simplify to $$y_1(t)={\left(1-e^{-t/2}\right)}^2,y_2(t)={\left(1-e^{-t/3}\right)}^3.$$ Both curves start at the origin with gradient $0$ (verified by differentiating and setting $t=0$), and both approach the asymptote $y=1$ as $t\to \infty$. To determine which lies higher, note that for $t>0$: since $\frac{t}{3}<\frac{t}{2}$, we have $e^{-t/3}>e^{-t/2}$, so $$1-e^{-t/3}<1-e^{-t/2}.$$ Call these values $a<b$ (both in $(0,1)$). We need to compare $a^3$ with $b^2$. For small $t$: $b\approx t/2$ and $a\approx t/3$, giving $b^2\approx t^2/4\gg t^3/27\approx a^3$, so $y_1>y_2$ near the origin. As $t\to \infty$ both $a,b\to 1$, and $1-b\approx e^{-t/2}\ll e^{-t/3}\approx 1-a$, so $b^2\to 1$ faster than $a^3\to 1$, confirming $y_1>y_2$ throughout. Therefore $y_1$ lies strictly above $y_2$ for all $t>0$, with both curves increasing from $(0,0)$, both concave (decreasing gradient after the origin), and both having the asymptote $y=1$.

Part (i).

To find the linear solution, substitute $y=mx+c$ into $\frac{dy}{dx}=y+x+1$. The left-hand side gives $m$, while the right-hand side gives $mx+c+x+1=(m+1)x+(c+1)$. Equating coefficients of $x$ and the constant: $m=m+1$ gives $m=-1$, and then $m=c+1$ gives $c=-2$. So the linear solution is $y=-x-2$.

At a stationary point of $y_3(x)$, we have $\frac{dy}{dx}=0$, so the equation gives $y+x+1=0$, i.e. $y=-x-1$. Hence all stationary points lie on the line $y=-x-1$.

Now, $y=-x-2$ is itself a solution curve. Two distinct solution curves cannot intersect (if they met at $(a,b)$ they would share the same gradient $b+a+1$ there, and by uniqueness they would be identical — a contradiction). Since the line $y=-x-2$ is a solution curve, the curve $y_3(x)$ cannot cross it. If $k<-2$, then $y_3(0)=k<-2$, so $y_3$ starts strictly below $y=-x-2$. The line of stationary points $y=-x-1$ lies strictly above $y=-x-2$ (it is parallel, shifted up by $1$), so $y_3$ can never reach it and has no stationary points.

To classify any stationary point that does occur, differentiate the equation: $$\frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}+1.$$ At a stationary point $\frac{dy}{dx}=0$, so $\frac{d^{2}y}{d{x}^2}=1>0$. Hence every stationary point is a local minimum.

For the substitution, let $Y=y+x$, so $\frac{dY}{dx}=\frac{dy}{dx}+1$. The equation becomes $\frac{dY}{dx}-1=Y+1$, i.e. $\frac{dY}{dx}=Y+2$. Separating variables: $$\int \frac{dY}{Y+2}=\int dx⟹\ln |Y+2|=x+c_0⟹Y+2=A{e}^x.$$ Back-substituting $Y=y+x$ gives $y=A{e}^x-x-2$. Applying $y_3(0)=k$ yields $A=k+2$, so $$y_3(x)=(k+2)e^x-x-2.$$ The three cases are: $k=0\Rightarrow y=2e^x-x-2$; $k=-2\Rightarrow y=-x-2$; $k=-3\Rightarrow y=-e^x-x-2$. All three curves share the asymptotic behaviour $y\approx -x-2$ as $x\to -\infty$. The curve $k=0$ has a minimum (on $y=-x-1$) and grows like $2e^x$ for large $x$; the line $k=-2$ is the separating solution; the curve $k=-3$ lies entirely below the line and decreases to $-\infty$.

Part (ii).

Substitute $y=mx+c$: the left-hand side is $m$, and the right-hand side expands as $(y-x{)}^2+4(y-x)+3$. With $y-x=(m-1)x+c$, for this to be constant in $x$ we need $m-1=0$, so $m=1$. Then $y-x=c$ and the equation gives $m=c^2+4c+3$, i.e. $1=c^2+4c+3$, so $c^2+4c+2=0$, giving $c=-2\pm \sqrt{2}$. The two linear solutions are $y=x-2+\sqrt{2}$ and $y=x-2-\sqrt{2}$.

At a stationary point, $\frac{dy}{dx}=0$, so $(y-x{)}^2+4(y-x)+3=0$, which factorises as $(y-x+1)(y-x+3)=0$. Hence stationary points lie on $y=x-1$ or $y=x-3$.

Between the two lines $y=x-1$ and $y=x-3$, we have $-3<y-x<-1$, so $(y-x+1)<0$ and $(y-x+3)>0$, giving $\frac{dy}{dx}<0$. Above both lines or below both lines, both factors carry the same sign and the gradient is positive.

Since $y_4(0)=-2$, the curve starts between the two straight-line solutions (which pass through $(0,-2\pm \sqrt{2})$). In this region the gradient is negative, so the curve is decreasing. It will eventually reach the lower stationary-point line $y=x-3$, where it has a local minimum, then increase toward the upper line $y=x-1$ where it has a local maximum, and so on. The sketch shows $y_4$ oscillating between the two stationary-point lines, with the two straight-line solutions acting as asymptotic boundaries above and below.

Part (i).

Substitute $y=(x-a)u$ into the equation. Differentiating gives$$\frac{dy}{dx}=(x-a)\frac{du}{dx}+u.$$Plugging into $(x-a)\frac{dy}{dx}=y-x$:$$(x-a)\left[(x-a)\frac{du}{dx}+u\right]=(x-a)u-x.$$The $u$ terms cancel on both sides, leaving$$(x-a{)}^2\frac{du}{dx}=-x,\text{i.e.}\frac{du}{dx}=\frac{-x}{(x-a{)}^2}.$$To integrate the right-hand side, write $-x=-(x-a)-a$, so$$\frac{du}{dx}=\frac{-(x-a)-a}{(x-a{)}^2}=-\frac{1}{x-a}-\frac{a}{(x-a{)}^2}.$$Integrating term by term:$$u=-\ln |x-a|+\frac{a}{x-a}+c.$$Since $u=\frac{y}{x-a}$, multiply through by $(x-a)$:$$\boxed{y=-(x-a)\ln |x-a|+a+c(x-a).}$$Part (ii).

The curve $C$ has equation $y=f(x)$. The condition says: for $t\ne 1$, the tangent to $C$ at $(t,f(t))$ passes through $(1,t)$.

The gradient of the tangent at $(t,f(t))$ is $f^{\prime }(t)$. But the gradient can also be computed from the two points $(t,f(t))$ and $(1,t)$:$$f^{\prime }(t)=\frac{f(t)-t}{t-1}.$$Rearranging: $(t-1)f^{\prime }(t)=f(t)-t$, which is exactly the differential equation from part (i) with $x\mapsto t$, $y\mapsto f(t)$, and $a=1$. The general solution is therefore$$f(t)=-(t-1)\ln |t-1|+1+c(t-1).$$Replacing $t$ with $x$: $f(x)=-(x-1)\ln |x-1|+1+c(x-1)$.

(a) Apply $f(0)=0$:$$0=-(-1)\ln 1+1+c(-1)=0+1-c⟹c=1.$$So $f(x)=-(x-1)\ln |x-1|+x$ for $x<1$.

Stationary points. From the original ODE, $f^{\prime }(x)=\frac{f(x)-x}{x-1}=-\ln |x-1|$. Setting $f^{\prime }(x)=0$ gives $|x-1|=1$, so $x=0$ or $x=2$. In the range $x<1$, the only stationary point is at $x=0$, where $f(0)=0$, giving the point $(0,0)$.

Behaviour near $x=1$. As $x\to 1^{-}$, $f^{\prime }(x)=-\ln |x-1|\to +\infty$, so the curve becomes vertical. Using the given hint $z\ln |z|\to 0$ as $z\to 0$, we get $f(x)\to 1$ as $x\to 1^{-}$. As $x\to -\infty$, $f^{\prime }(x)\to -\infty$.

Sketch for $x<1$: the curve passes through $(0,0)$ (a stationary point), rises steeply towards $(1,1)$ with a vertical tangent there, and falls away to the left with $f^{\prime }(x)\to -\infty$.

(b) Apply $f(2)=2$ to $f(x)=-(x-1)\ln |x-1|+1+c(x-1)$:$$2=-(1)\ln 1+1+c(1)=0+1+c⟹c=1.$$The formula is again $f(x)=-(x-1)\ln |x-1|+x$ and $f^{\prime }(x)=-\ln |x-1|$ for $x>1$. The stationary point from $|x-1|=1$ in this region is $x=2$, giving the point $(2,2)$. As $x\to 1^{+}$, $f^{\prime }(x)\to +\infty$ (vertical tangent, $f(x)\to 1^{+}$). As $x\to +\infty$, $f^{\prime }(x)\to -\infty$.

Sketch for $x>1$: starts near $(1,1)$ with a vertical tangent, rises to a maximum at $(2,2)$, then decreases with ever-steepening negative gradient.