Ordinary Differential Equations: Very Difficult Problems with Solutions

Solution:
Let the stone be at the point [tex](x, y)[/tex], [tex]0\lt x\le a[/tex]. The chain lies on the tangent to the trajectory at this point, and its other end is the point where this tangent meets the y-axis, i.e. [tex]\left(0, \ y-xy'\right)[/tex]. The distance between the two ends equals [tex]a[/tex]:
[tex]x^2+(xy')^2=a^2\ \Rightarrow\ y'=\pm\frac{\sqrt{a^2-x^2}}{x}[/tex]
As the end of the chain moves up, the stone moves up and towards the y-axis ([tex]y[/tex] increases while [tex]x[/tex] decreases), so we take the minus sign, with the initial condition [tex]y(a)=0[/tex]:
[tex]y'=-\frac{\sqrt{a^2-x^2}}{x},\qquad y(a)=0[/tex]
Integrating (for example with the substitution [tex]x=a\sin \theta[/tex]):
[tex]\int\frac{\sqrt{a^2-x^2}}{x}\,dx=\sqrt{a^2-x^2}-a\ln\frac{a+\sqrt{a^2-x^2}}{x}+C[/tex]
[tex]y=a\ln\frac{a+\sqrt{a^2-x^2}}{x}-\sqrt{a^2-x^2}+C,\qquad y(a)=0\ \Rightarrow\ C=0[/tex]
Hence the trajectory (a tractrix) is [tex]y=a\ln\frac{a+\sqrt{a^2-x^2}}{x}-\sqrt{a^2-x^2}[/tex], [tex]0\lt x\le a[/tex].

Solution:
Use polar coordinates along the curve: [tex]x=r\cos\varphi[/tex], [tex]y=r\sin \varphi[/tex], where [tex]r^2=x^2+y^2[/tex] and, since [tex]x\gt 0[/tex] near the point [tex](1, 1)[/tex], [tex]\varphi=\arctan\frac{y}{x}[/tex]. Differentiating with respect to [tex]x[/tex]:
[tex]rr'=x+yy',\qquad \varphi'=\frac{xy'-y}{x^2+y^2}\ \Rightarrow\ y-xy'=-r^2\varphi'[/tex]
The equation becomes
[tex]\frac{-r^2\varphi'}{rr'}=2\ \Rightarrow\ \frac{r'}{r}=-\frac{\varphi'}{2}\ \Rightarrow\ \ln r=-\frac{\varphi}{2}+C_1[/tex]
i.e. [tex]\ln\left(x^2+y^2\right)+\arctan\frac{y}{x}=C[/tex]. At the point [tex](1, 1)[/tex] we get [tex]C=\ln 2+\frac{\pi}{4}[/tex]. The initial condition is admissible: from [tex]y-xy'=2\left(x+yy'\right)[/tex] we get [tex]y'=\frac{y-2x}{x+2y}[/tex], so [tex]y'(1)=-\frac{1}{3}[/tex] and [tex]x+yy'=\frac{2}{3}\ne 0[/tex] at the initial point. Hence the solution is
[tex]\ln\left(x^2+y^2\right)+\arctan\frac{y}{x}=\ln 2+\frac{\pi}{4}[/tex]
This is the logarithmic spiral [tex]r=\sqrt{2}\,e^{\frac{1}{2}\left(\frac{\pi}{4}-\varphi\right)}[/tex].

Solution:
Collect the terms with [tex]y'[/tex]: [tex]y'\left(x-x^3\sin y\right)=2y[/tex]. The function [tex]y\equiv 0[/tex] is a solution. For [tex]y\ne 0[/tex] it is convenient to regard [tex]x[/tex] as a function of [tex]y[/tex]:
[tex]\frac{dx}{dy}=\frac{x-x^3\sin y}{2y}\ \Rightarrow\ \frac{dx}{dy}-\frac{1}{2y}\,x=-\frac{\sin y}{2y}\,x^3[/tex]
This is a Bernoulli equation for [tex]x(y)[/tex]. Put [tex]u=x^{-2}[/tex] ([tex]x\ne 0[/tex]); then [tex]\frac{du}{dy}=-2x^{-3}\frac{dx}{dy}[/tex] and we obtain a linear equation:
[tex]\frac{du}{dy}+\frac{u}{y}=\frac{\sin y}{y}\ \Rightarrow\ \frac{d}{dy}\left(yu\right)=\sin y\ \Rightarrow\ yu=C-\cos y[/tex]
Returning to [tex]x[/tex]: [tex]\frac{y}{x^2}=C-\cos y[/tex], so the general integral is
[tex]y=x^2(C-\cos y)[/tex]
(the solution [tex]y=0[/tex] is also contained in it, for [tex]C=1[/tex]). Check: differentiating [tex]y=x^2(C-\cos y)[/tex] gives
[tex]y'=2x(C-\cos y)+x^2y'\sin y=\frac{2y}{x}+x^2y'\sin y[/tex]
and multiplying by [tex]x[/tex] gives exactly [tex]xy'=2y+x^3y'\sin y[/tex].

Solution:
Write the equation as [tex]P\,dx+Q\,dy=0[/tex] with [tex]P=xy^2-y[/tex], [tex]Q=x^2y-3x[/tex]. Since [tex]P_y=2xy-1\ne Q_x=2xy-3[/tex], it is not exact. Look for an integrating factor [tex]\mu(z)[/tex], [tex]z=xy[/tex]. The condition [tex](\mu P)_y=(\mu Q)_x[/tex] gives
[tex]\mu'(z)\,xP+\mu P_y=\mu'(z)\,yQ+\mu Q_x\ \Rightarrow\ \frac{\mu'}{\mu}=\frac{Q_x-P_y}{xP-yQ}[/tex]
Here
[tex]Q_x-P_y=-2,\qquad xP-yQ=x^2y^2-xy-x^2y^2+3xy=2xy[/tex]
so [tex]\frac{\mu'}{\mu}=-\frac{1}{z}[/tex] depends only on [tex]z[/tex], and [tex]\mu=\frac{1}{z}=\frac{1}{xy}[/tex]. Multiplying by it (for [tex]xy\ne 0[/tex]) gives an exact equation and its potential:
[tex]\left(y-\frac{1}{x}\right)dx+\left(x-\frac{3}{y}\right)dy=0,\qquad F=xy-\ln|x|-3\ln|y|[/tex]
(indeed [tex]F_x=y-\frac{1}{x}[/tex], [tex]F_y=x-\frac{3}{y}[/tex]). Hence
[tex]xy-\ln|x|-3\ln|y|=C_1\ \Rightarrow\ \ln\left|xy^3\right|=xy-C_1\ \Rightarrow\ xy^3=Ce^{xy}[/tex]
The solution [tex]y=0[/tex], lost when multiplying by [tex]\mu[/tex], is contained in this family for [tex]C=0[/tex].

Solution:
This is a Riccati equation. Look for a particular solution of the form [tex]y_1=\frac{a}{x}[/tex]:
[tex]-\frac{a}{x^2}=\frac{a^2-2}{x^2}\ \Rightarrow\ a^2+a-2=0\ \Rightarrow\ a=1\text{ or }a=-2[/tex]
Take [tex]y_1=\frac{1}{x}[/tex] and substitute [tex]y=\frac{1}{x}+\frac{1}{v}[/tex]:
[tex]-\frac{1}{x^2}-\frac{v'}{v^2}=\frac{1}{x^2}+\frac{2}{xv}+\frac{1}{v^2}-\frac{2}{x^2}\ \Rightarrow\ v'+\frac{2}{x}\,v=-1[/tex]
This linear equation can be written as [tex]\left(x^2v\right)'=-x^2[/tex], so
[tex]x^2v=K-\frac{x^3}{3}\ \Rightarrow\ v=\frac{3K-x^3}{3x^2}\ \Rightarrow\ y=\frac{1}{x}+\frac{3x^2}{3K-x^3}[/tex]
Then
[tex]xy=\frac{3K+2x^3}{3K-x^3},\qquad 1-xy=\frac{-3x^3}{3K-x^3},\qquad 2+xy=\frac{9K}{3K-x^3}[/tex]
so [tex]\frac{1-xy}{2+xy}=-\frac{x^3}{3K}[/tex], and with [tex]C=-\frac{1}{3K}[/tex] we get
[tex]1-xy=Cx^3(2+xy)[/tex]
The particular solution [tex]y=\frac{1}{x}[/tex] corresponds to [tex]C=0[/tex], and [tex]y=-\frac{2}{x}[/tex] (the case [tex]K=0[/tex]) is obtained in the limit [tex]C\to\infty[/tex].

Solution:
The equation is homogeneous in [tex]y,y',y''[/tex]. Near [tex]x=1[/tex] we have [tex]y\gt 0[/tex], so we divide by [tex]y^2[/tex] and put [tex]z=\frac{y'}{y}[/tex]. Then [tex]\frac{y''}{y}=z'+z^2[/tex], [tex]z(1)=\frac{y'(1)}{y(1)}=1[/tex], and the equation becomes
[tex]x^2(z'+z^2)-2x^2z^2+xz=0\iff z'+\frac{z}{x}=z^2[/tex]
This is a Bernoulli equation. Near [tex]x=1[/tex] we have [tex]z\ne 0[/tex]; we put [tex]u=\frac{1}{z}[/tex], so that [tex]u'=-\frac{z'}{z^2}[/tex] and [tex]u(1)=1[/tex]:
[tex]u'-\frac{u}{x}=-1\iff\left(\frac{u}{x}\right)'=-\frac{1}{x}\Rightarrow u=x(K-\ln x)[/tex]
From [tex]u(1)=1[/tex] we get [tex]K=1[/tex], so [tex]\frac{y'}{y}=\frac{1}{x(1-\ln x)}[/tex]. Integrating (note that [tex]d(1-\ln x)=-\frac{dx}{x}[/tex]):
[tex]\ln y=-\ln|1-\ln x|+c[/tex]
The condition [tex]y(1)=1[/tex] gives [tex]c=0[/tex]. Therefore
[tex]y=\frac{1}{1-\ln x},\quad 0\lt x\lt e[/tex]
Check:
[tex]y'=\frac{1}{x(1-\ln x)^2},\quad y'(1)=1,\quad y''=\frac{1+\ln x}{x^2(1-\ln x)^3}[/tex]
[tex]x^2yy''-2x^2(y')^2+xyy'=\frac{(1+\ln x)-2+(1-\ln x)}{(1-\ln x)^4}=0[/tex]

Solution:
This is an Euler equation. We substitute [tex]x=e^{t}[/tex], i.e. [tex]t=\ln x\gt 0[/tex]; then (dots denote derivatives with respect to [tex]t[/tex]) [tex]xy'=\dot y[/tex], [tex]x^{2}y''=\ddot y-\dot y[/tex], and the equation becomes
[tex]\ddot y-2\dot y+y=\frac{e^{t}}{t}[/tex]
The characteristic equation [tex]\lambda^{2}-2\lambda+1=0[/tex] has the double root [tex]1[/tex], so the general solution of the homogeneous equation is
[tex]y_0=(C_1+C_2t)e^{t}[/tex]
The right-hand side is not a quasi-polynomial, so we use variation of constants: [tex]y=C_1(t)e^{t}+C_2(t)te^{t}[/tex], where
[tex]\begin{cases}C_1'e^{t}+C_2'te^{t}=0\\ C_1'e^{t}+C_2'(t+1)e^{t}=\frac{e^{t}}{t}\end{cases}\Rightarrow\ C_2'=\frac{1}{t},\quad C_1'=-1[/tex]
Hence [tex]C_1=-t+\tilde C_1[/tex], [tex]C_2=\ln t+\tilde C_2[/tex] and
[tex]y=\left(\tilde C_1+\tilde C_2t+t\ln t-t\right)e^{t}=\left(C_1+C_2t+t\ln t\right)e^{t}[/tex]
(the term [tex]-te^{t}[/tex] is absorbed by [tex]\tilde C_2te^{t}[/tex]). Returning to [tex]x[/tex] ([tex]e^{t}=x[/tex], [tex]t=\ln x[/tex]), we obtain
[tex]y=x\left(C_1+C_2\ln x+\ln x\,\ln(\ln x)\right)[/tex]

Solution:
We eliminate [tex]y[/tex]: we differentiate the first equation and use the second one.
[tex]x''=y'+\frac{2\tan t}{\cos^{2}t}=-x+\tan t+\frac{2\sin t}{\cos^{3}t}\ \Rightarrow\ x''+x=\tan t+\frac{2\sin t}{\cos^{3}t}[/tex]
The homogeneous equation gives [tex]x_0=C_1\cos t+C_2\sin t[/tex]. Since [tex](\tan t)'=\frac{1}{\cos^{2}t}[/tex] and [tex](\tan t)''=\frac{2\sin t}{\cos^{3}t}[/tex], the function [tex]X=\tan t[/tex] is a particular solution:
[tex]X''+X=\frac{2\sin t}{\cos^{3}t}+\tan t\ \Rightarrow\ x=C_1\cos t+C_2\sin t+\tan t[/tex]
From the first equation, using [tex]\frac{1}{\cos^{2}t}-\tan ^{2}t=1[/tex]:
[tex]y=x'-\tan ^{2}t+1=-C_1\sin t+C_2\cos t+\frac{1}{\cos^{2}t}-\tan ^{2}t+1=-C_1\sin t+C_2\cos t+2[/tex]
Check of the second equation: [tex]y'=-C_1\cos t-C_2\sin t=-x+\tan t[/tex].

Solution:
This is an Euler equation with respect to [tex]x+1[/tex]. For [tex]x\gt -1[/tex] we put [tex]x+1=e^{t}[/tex]; then [tex](x+1)y'=\dot y[/tex], [tex](x+1)^{2}y''=\ddot y-\dot y[/tex] (dots denote derivatives with respect to [tex]t[/tex]), and the equation becomes
[tex]\ddot y-2\dot y+y=-1[/tex]
The characteristic root [tex]1[/tex] is double and the constant [tex]-1[/tex] is a particular solution, so
[tex]y=(C_1+C_2t)e^{t}-1=C_1(x+1)+C_2(x+1)\ln(x+1)-1[/tex]
Since [tex]y'=C_1+C_2\left(\ln(x+1)+1\right)[/tex], the initial conditions give
[tex]y(0)=C_1-1=0,\quad y'(0)=C_1+C_2=0\ \Rightarrow\ C_1=1,\ C_2=-1,\quad y=x-(x+1)\ln(x+1)[/tex]
For [tex]|x|\lt 1[/tex] we have [tex]\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{n}}{n}[/tex], hence
[tex](1+x)\ln(1+x)=x+\sum_{n=2}^{\infty}\frac{(-1)^{n}x^{n}}{n(n-1)}\ \Rightarrow\ y=\sum_{n=2}^{\infty}\frac{(-1)^{n+1}x^{n}}{n(n-1)}=-\frac{x^{2}}{2}+\frac{x^{3}}{6}-\frac{x^{4}}{12}+\ldots[/tex]
By the ratio test, with [tex]a_n=\frac{(-1)^{n+1}}{n(n-1)}[/tex]:
[tex]\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{n-1}{n+1}\to 1\ \Rightarrow\ R=1[/tex]
This agrees with the theory: after division by [tex](x+1)^{2}[/tex] the coefficients are analytic for [tex]|x|\lt 1[/tex], so [tex]R\ge 1[/tex]; and [tex]R\gt 1[/tex] is impossible, since then [tex]y''[/tex] would stay bounded near [tex]x=-1[/tex], while [tex]y''=-\frac{1}{x+1}\to-\infty[/tex] as [tex]x\to -1^{+}[/tex].

Solution:
The condition can be written as
[tex]e^{-f(x)}f'(x)=e^{-g(x)}g'(x)\iff\left(e^{-f(x)}\right)'=\left(e^{-g(x)}\right)'[/tex]
Hence [tex]e^{-f(x)}-e^{-g(x)}=K[/tex] is a constant on [tex]\mathbb{R}[/tex]. Comparing [tex]x=0[/tex] and [tex]x=2018[/tex]:
[tex]e^{-1}-e^{-g(0)}=e^{-f(2018)}-e^{-1}\ \Rightarrow\ e^{-f(2018)}=\frac{2}{e}-e^{-g(0)}\lt\frac{2}{e}[/tex]
Therefore, for all admissible [tex]f[/tex] and [tex]g[/tex],
[tex]f(2018)\gt -\ln\frac{2}{e}=1-\ln 2[/tex]
The bound cannot be improved. Take [tex]g(x)=1+\lambda(2018-x)[/tex] with [tex]\lambda\gt 0[/tex] (then [tex]g'=-\lambda\ne 0[/tex], [tex]g(2018)=1[/tex], [tex]g(0)\gt 1[/tex]) and define [tex]f[/tex] by
[tex]e^{-f(x)}=e^{-g(x)}+e^{-1}-e^{-g(0)}[/tex]
The right-hand side is positive because [tex]e^{-1}-e^{-g(0)}\gt 0[/tex]; moreover [tex]f(0)=1[/tex], and differentiation gives [tex]e^{-f}f'=e^{-g}g'[/tex], which is the required relation. For these functions
[tex]f(2018)=-\ln\left(\frac{2}{e}-e^{-1-2018\lambda}\right)\to 1-\ln 2\quad (\lambda\to+\infty)[/tex]
So no constant larger than [tex]1-\ln 2[/tex] works, and [tex]c=1-\ln 2[/tex].

Solution:
For [tex]y\ne 0[/tex], separation of variables gives
[tex]\frac{dy}{3y^{2/3}}=dx\ \Rightarrow\ y^{1/3}=x+C\ \Rightarrow\ y=(x+C)^{3}[/tex]
Also [tex]y\equiv 0[/tex] is a solution. Both [tex]y\equiv 0[/tex] and [tex]y=(x-2)^{3}[/tex] satisfy [tex]y(2)=0[/tex], but there are many more solutions. For any numbers [tex]a\le 2\le b[/tex] define
[tex]y_{a,b}(x)=\begin{cases}(x-a)^{3}, & x\lt a\\ 0, & a\le x\le b\\ (x-b)^{3}, & x\gt b\end{cases}[/tex]
For [tex]x\gt b[/tex]:
[tex]y'=3(x-b)^{2}=3\sqrt[3]{(x-b)^{6}}=3\sqrt[3]{y^{2}}[/tex]
For [tex]x\lt a[/tex] the same computation works, since [tex]\sqrt[3]{(x-a)^{6}}=(x-a)^{2}[/tex] although [tex]x-a\lt 0[/tex]. For [tex]a\lt x\lt b[/tex] both sides equal [tex]0[/tex]. At [tex]x=a[/tex] and [tex]x=b[/tex] the function is continuous and both one-sided derivatives are [tex]0[/tex]; for example,
[tex]\lim_{x\to a^{-}}\frac{y(x)-y(a)}{x-a}=\lim_{x\to a^{-}}(x-a)^{2}=0[/tex]
so [tex]y'(a)=y'(b)=0=3\sqrt[3]{y^{2}}[/tex] at these points. Hence every [tex]y_{a,b}[/tex] is a solution on [tex]\mathbb{R}[/tex] with [tex]y(2)=0[/tex], and different pairs [tex](a, b)[/tex] give different functions. Therefore the problem has infinitely many solutions.
This does not contradict the existence and uniqueness theorem: [tex]f(y)=3y^{2/3}[/tex] is continuous but does not satisfy a Lipschitz condition near [tex]y=0[/tex], since
[tex]\frac{|f(y)-f(0)|}{|y|}=3|y|^{-1/3}\to\infty\quad (y\to 0)[/tex]

Solution:
Consider the function [tex]E(x)=f(x)^2+f'(x)^2[/tex]; it is differentiable because [tex]f[/tex] is twice differentiable. By the given equation,
[tex]E'(x)=2f'(x)\left(f(x)+f''(x)\right)=-2x\,g(x)\,f'(x)^2[/tex]
Since [tex]g\gt 0[/tex] and [tex]f'(x)^2\ge 0[/tex], we have [tex]E'(x)\ge 0[/tex] for [tex]x\le 0[/tex] and [tex]E'(x)\le 0[/tex] for [tex]x\ge 0[/tex]. Hence [tex]E[/tex] is non-decreasing on [tex]\left(-\infty, 0\right][/tex] and non-increasing on [tex]\left[0, +\infty\right)[/tex], so [tex]E(x)\le E(0)[/tex] for every [tex]x[/tex]. Consequently
[tex]f(x)^2\le E(x)\le E(0)=f(0)^2+f'(0)^2\ \Rightarrow\ |f(x)|\le\sqrt{f(0)^2+f'(0)^2}[/tex]
for all [tex]x\in\mathbb{R}[/tex], so [tex]|f|[/tex] is bounded.

Solution:
First we check that both functions are differentiable on [tex](0, 2)[/tex]. For [tex]x\ne 1[/tex] this is clear. At [tex]x=1[/tex], since [tex]\alpha\gt 1[/tex],
[tex]\lim_{x\to 1^{-}}\frac{y_1(x)-y_1(1)}{x-1}=-\lim_{x\to 1^{-}}(1-x)^{\alpha-1}=0[/tex]
and the right derivative is [tex]0[/tex] because [tex]y_1=0[/tex] on [tex]\left[1, 2\right)[/tex]; so [tex]y_1'(1)=0[/tex]. In the same way [tex]y_2'(1)=0[/tex]. Thus
[tex]y_1'(x)=\begin{cases}-\alpha(1-x)^{\alpha-1}, & 0\lt x\lt 1\\ 0, & 1\le x\lt 2\end{cases}\qquad y_2'(x)=\begin{cases}0, & 0\lt x\le 1\\ \alpha(x-1)^{\alpha-1}, & 1\lt x\lt 2\end{cases}[/tex]
The Wronskian is [tex]W=y_1y_2'-y_1'y_2[/tex]. On [tex](0, 1)[/tex] we have [tex]y_2=y_2'=0[/tex], and on [tex]\left[1, 2\right)[/tex] we have [tex]y_1=y_1'=0[/tex]; in both cases [tex]W=0[/tex]. Hence [tex]W\equiv 0[/tex] on [tex](0, 2)[/tex].
Linear independence: suppose that [tex]c_1y_1(x)+c_2y_2(x)=0[/tex] for all [tex]x\in(0, 2)[/tex]. Substituting [tex]x=\frac{1}{2}[/tex] and [tex]x=\frac{3}{2}[/tex]:
[tex]c_1\left(\frac{1}{2}\right)^{\alpha}=0\Rightarrow c_1=0,\qquad c_2\left(\frac{1}{2}\right)^{\alpha}=0\Rightarrow c_2=0[/tex]
Therefore [tex]y_1[/tex] and [tex]y_2[/tex] are linearly independent on [tex](0, 2)[/tex], although [tex]W\equiv 0[/tex].
Remark. This does not contradict the theorem that an identically zero Wronskian implies linear dependence: that theorem holds only for solutions of one linear homogeneous equation [tex]y''+p(x)y'+q(x)y=0[/tex] with continuous coefficients. Here [tex]y_1[/tex] cannot be such a solution: [tex]y_1(1)=y_1'(1)=0[/tex], so by the uniqueness theorem it would vanish identically, but [tex]y_1\not\equiv 0[/tex].