Differential Equations

Part (i).

The equation $u^2+2u\sinh x-1=0$ is a quadratic in $u$. Using the quadratic formula:$$u=\frac{-2\sinh x\pm \sqrt{4{\sinh }^{2}x+4}}{2}=-\sinh x\pm \sqrt{{\sinh }^{2}x+1}=-\sinh x\pm \cosh x.$$Since $\cosh x>0$, this gives $u=e^{-x}$ or $u=-e^x$.

Now treat the differential equation: replace $u$ by $\frac{\mathrm{d}y}{\mathrm{d}x}$. The two cases give:$$\frac{\mathrm{d}y}{\mathrm{d}x}=e^{-x}\Rightarrow y=-e^{-x}+C_1,$$$$\frac{\mathrm{d}y}{\mathrm{d}x}=-e^x\Rightarrow y=-e^x+C_2.$$Applying $y=0$ and $\frac{\mathrm{d}y}{\mathrm{d}x}>0$ at $x=0$: the second branch gives $\frac{\mathrm{d}y}{\mathrm{d}x}{|}_{x=0}=-1<0$, so it is rejected. The first branch gives $\frac{\mathrm{d}y}{\mathrm{d}x}{|}_{x=0}=1>0$ and $0=-1+C_1$, so $C_1=1$. The required solution is:$$\boxed{y=1-e^{-x}}.$$Part (ii).

Divide through by $\sinh y$ (justified because we seek a non-trivial solution, so $y\ne 0$ generically):$${\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2+\frac{2}{\sinh y}\frac{\mathrm{d}y}{\mathrm{d}x}-1=0.$$This is the same quadratic form as Part (i) but with $\sinh y$ replaced by $\frac{1}{\sinh y}$. Setting $v=\frac{\mathrm{d}y}{\mathrm{d}x}$, the quadratic $v^2+\frac{2v}{\sinh y}-1=0$ gives:$$v=\frac{-1\pm \cosh y}{\sinh y}.$$Hence $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\cosh y-1}{\sinh y}$ or $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-(\cosh y+1)}{\sinh y}$, i.e.:$$\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{\sinh y}{\cosh y-1}\text{or}\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{-\sinh y}{\cosh y+1}.$$Integrate each: the first gives $x=\ln (\cosh y-1)+C_1$, and the second $x=-\ln (\cosh y+1)+C_2$. Only the first branch can satisfy $y=0$ at $x=0$ (the second gives $x=-\ln 2+C_2$ at $y=0$; with $x=0$ that puts $C_2=\ln 2$, so $\cosh y+1=2e^{-x}$, which holds also). Actually applying $x=0,y=0$ to the first: $0=\ln (\cosh 0-1)+C_1=\ln 0+C_1$, which is $-\infty$. So the first branch does not satisfy the initial condition. The second branch with $C_2=\ln 2$ gives:$$x=\ln 2-\ln (\cosh y+1)=\ln \frac{2}{\cosh y+1}.$$Rearranging:$$e^x=\frac{2}{\cosh y+1}\Rightarrow \cosh y+1=2e^{-x}\Rightarrow \boxed{\cosh y=2e^{-x}-1}.$$Asymptotes.

As $x\to -\infty$, $e^{-x}\to +\infty$, so $\cosh y\to +\infty$, meaning $y\to \pm \infty$ (since $\cosh$ is even). In that regime $\cosh y\approx \frac{e^{|y|}}{2}$, so:$$\frac{e^{|y|}}{2}\approx 2e^{-x}\Rightarrow e^{|y|}\approx 4e^{-x}\Rightarrow |y|\approx -x+\ln 4.$$For $y>0$: $y\to -x+\ln 4$. For $y<0$: $y\to x-\ln 4$, equivalently $y=-(-x+\ln 4)$. These are the asymptotes $y=\pm (-x+\ln 4)$, as required. Note that for $x>0$ we need $2e^{-x}>1$, i.e. $e^{-x}>\frac{1}{2}$, i.e. $x<\ln 2$, so the curve only exists for $x<\ln 2$.

This question explores Clairaut-type and related ODEs, where treating $p=\frac{\mathrm{d}y}{\mathrm{d}x}$ as the variable and inverting the roles of $x$ and $p$ turns a nonlinear problem into a linear one.

Part (i).

Differentiate $y=p^2+2xp$ with respect to $x$:$$p=2p\frac{\mathrm{d}p}{\mathrm{d}x}+2p+2x\frac{\mathrm{d}p}{\mathrm{d}x}.$$Collect and rearrange:$$p-2p=(2p+2x)\frac{\mathrm{d}p}{\mathrm{d}x}⟹-p=2(p+x)\frac{\mathrm{d}p}{\mathrm{d}x}.$$Divide through by $\frac{\mathrm{d}p}{\mathrm{d}x}$ and invert to get $\frac{\mathrm{d}x}{\mathrm{d}p}$:$$\frac{\mathrm{d}x}{\mathrm{d}p}=-\frac{2(p+x)}{p}=-2-\frac{2x}{p},$$as required.

This is a linear first-order ODE in $x(p)$:$$\frac{\mathrm{d}x}{\mathrm{d}p}+\frac{2}{p}x=-2.$$The integrating factor is $e^{\int (2/p)\mathrm{d}p}=p^2$. Multiplying through:$$\frac{\mathrm{d}}{\mathrm{d}p}(p^{2}x)=-2p^2.$$Integrate:$$p^{2}x=-\frac{2p^3}{3}+C⟹x=-\frac{2}{3}p+A{p}^{-2},$$where $A=C$ is an arbitrary constant.

To find $y$, apply the initial condition $p=-3$, $x=2$:$$2=-\frac{2}{3}(-3)+A(-3{)}^{-2}=2+\frac{A}{9}⟹A=0.$$So $x=-\frac{2}{3}p$, giving $p=-\frac{3}{2}x$. Substitute into the original equation:$$y=p^2+2xp=\frac{9}{4}{x}^2+2x\left(-\frac{3}{2}x\right)=\frac{9}{4}{x}^2-3x^2=-\frac{3}{4}{x}^2.$$Part (ii).

Differentiate $y=2xp+p\ln p$ with respect to $x$:$$p=2p+2x\frac{\mathrm{d}p}{\mathrm{d}x}+(\ln p+1)\frac{\mathrm{d}p}{\mathrm{d}x}.$$Rearrange:$$-p=(2x+\ln p+1)\frac{\mathrm{d}p}{\mathrm{d}x}.$$Invert:$$\frac{\mathrm{d}x}{\mathrm{d}p}=-\frac{2x+\ln p+1}{p}=-\frac{2}{p}x-\frac{\ln p+1}{p}.$$This is again linear with integrating factor $p^2$:$$\frac{\mathrm{d}}{\mathrm{d}p}(p^{2}x)=-p(\ln p+1).$$Integrate the right-hand side using integration by parts ($\int p\ln p\mathrm{d}p=\frac{p^2}{2}\ln p-\frac{p^2}{4}$):$$p^{2}x=-\frac{p^2}{2}\ln p+\frac{p^2}{4}-\frac{p^2}{2}+B=-\frac{p^2}{2}\ln p-\frac{p^2}{4}+B.$$Divide by $p^2$:$$x=-\frac{1}{2}\ln p-\frac{1}{4}+B{p}^{-2}.$$Apply $p=1$, $x=-\frac{1}{4}$:$$-\frac{1}{4}=0-\frac{1}{4}+B⟹B=0.$$So $x=-\frac{1}{2}\ln p-\frac{1}{4}$, giving $\ln p=-2x-\frac{1}{2}$, hence $p=e^{-2x-1/2}$.

Substitute into $y=2xp+p\ln p$:$$y=p(2x+\ln p)=e^{-2x-1/2}\left(2x+\left(-2x-\frac{1}{2}\right)\right)=e^{-2x-1/2}\cdot \left(-\frac{1}{2}\right)=-\frac{1}{2}{e}^{-2x-1/2}.$$