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2025 Paper 2 Q7
D: 1500.0 B: 1500.0
differential equation second order nonlinear ODE reduction of order separation of variables particle motion initial value problem long-term behaviour

The differential equation \[\frac{d^2x}{dt^2} = 2x\frac{dx}{dt}\] describes the motion of a particle with position \(x(t)\) at time \(t\). At \(t = 0\), \(x = a\), where \(a > 0\).

  1. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2\) when \(t = 0\). What happens to the particle as \(t\) increases from 0?
  2. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2 + p\) when \(t = 0\), where \(p > 0\). What happens to the particle as \(t\) increases from 0?
  3. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2 - q^2\) when \(t = 0\), where \(q > 0\). What happens to the particle as \(t\) increases from 0? Give conditions on \(a\) and \(q\) for the different cases which arise.

Solution: Let \(v = \frac{\d x}{\d t}\) and notice that \(\frac{\d}{\d t} \left ( \frac{\d x}{\d t} \right) = \frac{\d }{\d x} \left ( v \right) \frac{\d x}{\d t} = v \frac{\d v}{\d x}\). Also notice that: \begin{align*} && v \frac{\d v}{\d x} &= 2x v \\ \Rightarrow && \frac{\d v}{\d x} &= 2x \\ \Rightarrow && v &= x^2 + C \\ \Rightarrow && \frac{\d x}{\d t} &= x^2 + C \\ \end{align*}

  1. When \(t = 0, \frac{\d x}{\d t} = a^2\) so \(C = 0\), therefore \(\frac{\d x}{\d t} = x^2 \Rightarrow t = -x^{-1} + k\) and so \(k = a^{-1}\) and \(x = \frac{a}{1-at}\). As \(t\) increases from \(0\) the particle heads to infinity at an increasing rate, `reaching' infinity around \(t=\frac{1}{a}\)
  2. When \(t = 0, \frac{\d x}{\d t} = a^2 + p\) so \(C = p\). Therefore \(\frac{\d x}{\d t} = x^2 + p \Rightarrow t = \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) + c\). When \(c = - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right)\), so \begin{align*} && t &= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right) \\ &&&= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} \right) \\ \Rightarrow && \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} &= \tan (\sqrt{p} t) \\ \Leftrightarrow && \sqrt{p}(x-a) &= \tan (\sqrt{p} t)(\sqrt{p}-ax) \\ \Leftrightarrow && x(\sqrt{p}+a\tan (\sqrt{p} t)) &= \sqrt{p} (\tan(\sqrt{p}t) + a) \\ \Leftrightarrow && x &= \frac{\sqrt{p} (\tan(\sqrt{p}t) + a)}{\sqrt{p}+a\tan (\sqrt{p} t)} \end{align*} The particle heads to \(\frac{\sqrt{p}}{a}\).
  3. When \(t = 0, \frac{\d x}{\d t} = a^2-q^2\) so \(C = -q^2\). Therefore \begin{align*} && \frac{\d x}{\d t} &= x^2 -q^2 \\ \Rightarrow && \int \d t &= \int \frac{1}{(x-q)(x+q)} \d x \\ &&&= \frac{1}{2q} \int \left ( \frac{1}{x-q}- \frac{1}{x+q} \right )\d x \\ &&&= \frac{1}{2q} \left ( \ln (x-q) - \ln(x+q) \right) \\ &&&= \frac{1}{2q} \ln \left ( \frac{x-q}{x+q} \right)\\ \Rightarrow && \frac{x-q}{x+q} &= Ae^{2qt} \\ \underbrace{\Rightarrow}_{t= 0} && A &= \frac{a-q}{a+q} \\ \Rightarrow && x-q &= \frac{a-q}{a+q}e^{2qt}(x+q) \\ \Leftrightarrow && x\left (1-\frac{a-q}{a+q}e^{2qt} \right) &= q\left (1 + \frac{a-q}{a+q}e^{2qt} \right) \\ \Leftrightarrow && x &= q \frac{1 + \frac{a-q}{a+q}e^{2qt}}{1-\frac{a-q}{a+q}e^{2qt}} \end{align*}

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2020 Paper 3 Q7
D: 1500.0 B: 1500.0
second order differential equations integrating factor first order differential equations substitution initial value problem differential equations

  1. Given that the variables \(x\), \(y\) and \(u\) are connected by the differential equations \[ \frac{\mathrm{d}u}{\mathrm{d}x} + \mathrm{f}(x)u = \mathrm{h}(x) \quad \text{and} \quad \frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{g}(x)y = u, \] show that \[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (\mathrm{g}(x) + \mathrm{f}(x))\frac{\mathrm{d}y}{\mathrm{d}x} + (\mathrm{g}'(x) + \mathrm{f}(x)\mathrm{g}(x))y = \mathrm{h}(x). \tag{1} \]
  2. Given that the differential equation \[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(1 + \frac{4}{x}\right)\frac{\mathrm{d}y}{\mathrm{d}x} + \left(\frac{2}{x} + \frac{2}{x^2}\right)y = 4x + 12 \tag{2} \] can be written in the same form as (1), find a first order differential equation which is satisfied by \(\mathrm{g}(x)\). If \(\mathrm{g}(x) = kx^n\), find a possible value of \(n\) and the corresponding value of \(k\). Hence find a solution of (2) with \(y = 5\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3\) at \(x = 1\).

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2014 Paper 2 Q5
D: 1600.0 B: 1486.1
differential equation first order homogeneous ODE substitution y=xu translation of axes quadratic in x and y initial value problem

Given that \(y=xu\), where \(u\) is a function of \(x\), write down an expression for \(\dfrac {\d y}{\d x}\).

  1. Use the substitution \(y=xu\) to solve \[ \frac {\d y}{\d x} = \frac {2y+x}{y-2x} \] given that the solution curve passes through the point \((1,1)\). Give your answer in the form of a quadratic in \(x\) and \(y\).
  2. Using the substitutions \(x=X+a\) and \(y=Y+b\) for appropriate values of \(a\) and \(b\), or otherwise, solve \[ \frac {\d y}{\d x} = \frac {x-2y-4} {2x+y-3}\,, \] given that the solution curve passes through the point \((1,1)\).

Solution: \begin{align*} \frac{\d y}{\d x} &= \frac{\d }{\d x} \l y \r \\ &= \frac{\d }{\d x} \l xu \r \\ &\underbrace{=}_{\text{product rule}} \frac{\d}{\d x} \l x \r u + x \frac{\d}{\d x} \l u \r \\ &= u + x \frac{\d u}{\d x} \end{align*} \begin{questionparts} \item \begin{align*} && \frac{\d y}{\d x} &= \frac{2y + x}{y - 2x} \\ && u + x \frac{\d u}{\d x} &= \frac{2u + 1}{u - 2} \\ && x \frac{\d u}{\d x} &= \frac{2u-1-u^2+2u}{u-2} \\ \Rightarrow && \int \frac{2-u}{u^2-4u+1} \d u &= \int \frac{1}{x} \d x \\ && \int \frac{2-u}{(u-2)^2-5} \d u &= \int \frac1x \d x \\ && -\frac12\ln| (u-2)^2 - 5| &= \ln x + C \\ (x,y) = (1,1): && - \ln 2 &= C \\ \Rightarrow && \ln x^2 &= \ln 4 - \ln |5 - (u-2)^2| \\ \Rightarrow && x^2 &= \frac{4}{5- (u-2)^2} \\ \Rightarrow && 4 & = x^2(5 - (\frac{y}{x} - 2)^2) \\ &&&= 5x^2 - (y-2x)^2 \\ &&&= x^2+4xy-y^2 \end{align*} \item It would be convienient if \(x-2y -4 = X-2Y\) and \(2x+y-3 = 2X+Y\), ie \(a-2b = 4\) and \(2a+b = 3\), ie \(a = 2, b = -1\). Now our differential equation is: \begin{align*} && \frac{\d Y}{\d X} &= \frac{X - 2Y}{2X+Y} \\ && \frac{\d X}{\d Y} &= \frac{2X + Y}{X-2Y} \end{align*} This is the same differential equation we have already solved, just with the roles of \(x\) and \(y\) interchanged with \(Y\) and \(X\) and with the point \((0,3)\) being on the curve, ie: \(Y^2 + 4XY-X^2 = c\) and \(c = 9\), therefore our equation is: \[ (y-1)^2 + 4(y-1)(x+2)-(x+2)^2 = 9\]

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2012 Paper 3 Q1
D: 1700.0 B: 1500.0
differential equation second order substitution chain rule product rule initial value problem nonlinear ODE reduction of order

Given that \(\displaystyle z = y^n \left( \frac{\d y}{\d x}\right)^{\!2}\), show that \[ \frac{\d z}{\d x} = y^{n-1} \frac{\d y}{\d x} \left( n \left(\frac{\d y}{\d x}\right)^{\!2} + 2y \frac{\d^2y}{\d x^2}\right) . \]

  1. Use the above result to show that the solution to the equation \[ \left(\frac{\d y}{\d x}\right)^{\!2} + 2y \frac{\d^2y}{\d x^2} = \sqrt y \ \ \ \ \ \ \ \ \ \ (y>0) \] that satisfies \(y=1\) and \(\dfrac{\d y}{\d x}=0\) when \(x=0\) is \(y= \big ( \frac38 x^2+1\big)^{\frac23}\).
  2. Find the solution to the equation \[ \left(\frac{\d y}{\d x}\right)^{\!2} -y \frac{\d^2y}{\d x^2} + y^2=0 \] that satisfies \(y=1\) and \(\dfrac{\d y}{\d x}=0\) when \(x=0\).

Solution: \begin{align*} &&z &= y^n \left( \frac{\d y}{\d x}\right)^{2} \\ \Rightarrow && \frac{\d z}{\d x} &= ny^{n-1}\left( \frac{\d y}{\d x}\right)^{3} + y^{n} \cdot 2 \left( \frac{\d y}{\d x}\right) \left( \frac{\d^2 y}{\d x^2}\right) \\ &&&= y^{n-1} \left( \frac{\d y}{\d x}\right) \left (n \left( \frac{\d y}{\d x}\right)^2 + 2y \frac{\d^2 y}{\d x^2} \right) \end{align*}

  1. Let \(z = y (y')^2\), then \begin{align*} && \frac{\d z}{\d x} &= y' \sqrt{y} \\ &&&= \sqrt{z} \\ \Rightarrow && \int z^{-1/2} \d z &= x+C \\ \Rightarrow && 2\sqrt{z} &= x + C \\ x = 0, z=0: && C &= 0 \\ \Rightarrow && y(y')^2 &= \frac14 x^2 \\ \Rightarrow &&\sqrt{y} \frac{\d y}{\d x} &= \frac{1}{2}x\\ \Rightarrow && \int \sqrt{y} \d y &= \int \frac{1}{2}x\d x \\ \Rightarrow && \frac23y^{3/2} &=\frac14x^2 + K \\ x = 0, y = 1: && K &= \frac23 \\ \Rightarrow && y &= \left (\frac38 x^2 + 1 \right)^{2/3} \end{align*}
  2. Let \(z = y^{-2} (y')^2\) \begin{align*} && \frac{\d z}{\d x} &= y^{-3} \frac{\d y}{\d x} \left (-2 \left( \frac{\d y}{\d x}\right) + 2y \frac{\d^2 y}{\d x^2} \right) \\ &&&= y^{-3} \frac{\d y}{\d x} 2y^2 \\ &&&= 2y^{-1}(y') = 2 \sqrt{z} \\ \Rightarrow && 2\sqrt{z} &= 2x + C \\ x = 0, z = 0: && C&= 0 \\ \Rightarrow && z &= x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= xy \\ \Rightarrow && \ln |y| &= \frac12 x^2 + K \\ x =0 , y =1; && K &= 0 \\ \Rightarrow && y &= e^{\frac12 x^2} \end{align*}

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2009 Paper 3 Q10
D: 1700.0 B: 1484.0
second order differential equation SHM elastic spring Hooke's law energy conservation particle on spring trigonometric initial value problem compression

A light spring is fixed at its lower end and its axis is vertical. When a certain particle \(P\) rests on the top of the spring, the compression is \(d\). When, instead, \(P\) is dropped onto the top of the spring from a height \(h\) above it, the compression at time \(t\) after \(P\) hits the top of the spring is \(x\). Obtain a second-order differential equation relating \(x\) and \(t\) for \(0\le t \le T\), where \(T\) is the time at which \(P\) first loses contact with the spring. Find the solution of this equation in the form \[ x= A + B\cos (\omega t) + C\sin(\omega t)\,, \] where the constants \(A\), \(B\), \(C\) and \(\omega\) are to be given in terms of \(d\), \(g\) and \(h\) as appropriate. Show that \[ T = \sqrt{d/g\;} \left (2 \pi - 2 \arctan \sqrt{2h/d\;}\;\right)\,. \]

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2008 Paper 2 Q7
D: 1600.0 B: 1472.0
differential equation substitution Bernoulli equation separation of variables initial value problem conjecture pattern recognition

  1. By writing \(y=u{(1+x^2)\vphantom{\dot A}}^{\frac12}\), where \(u\) is a function of \(x\), find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = xy + \frac x {1+x^2} \] for which \(y=1\) when \(x=0\).
  2. Find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^2y + \frac {x^2 } {1+x^3} \] for which \(y=1\) when \(x=0\).
  3. Give, without proof, a conjecture for the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^{n-1}y + \frac {x^{n-1} } {1+x^n} \] for which \(y=1\) when \(x=0\), where \(n\) is an integer greater than 1.

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2005 Paper 2 Q8
D: 1600.0 B: 1516.0
differential equation separable ODE integration partial fractions curve sketching asymptotes t-substitution initial value problem

For \(x \ge 0\) the curve \(C\) is defined by $$ {\frac{\d y} {\d x}} = \frac{x^3y^2}{(1 + x^2)^{5/2}} $$ with \(y = 1\) when \(x=0\,\). Show that \[ \frac 1 y = \frac {2+3x^2}{3(1+x^2)^{3/2}} +\frac13 \] and hence that for large positive \(x\) $$ y \approx 3 - \frac 9 x\;. $$ Draw a sketch of \(C\). On a separate diagram, draw a sketch of the two curves defined for \(x \ge 0\) by $$ \frac {\d z} {\d x} = \frac{x^3z^3}{2(1 + x^2)^{5/2}} $$ with \(z = 1\) at \(x=0\) on one curve, and \(z = -1\) at \(x=0\) on the other.

Solution: \begin{align*} && {\frac{\d y} {\d x}} &= \frac{x^3y^2}{(1 + x^2)^{5/2}} \\ \Rightarrow &&\int \frac{1}{y^2} \d y &= \int \frac{x^3}{(1+x^2)^{5/2}} \d x \\ \Rightarrow && -\frac1y &= \int \frac{x^3+x-x}{(1+x^2)^{5/2}} \d x \\ &&&= \int \left ( \frac{x}{(1+x^2)^{3/2}}-\frac{x}{(1+x^2)^{5/2}} \right) \d x \\ &&&= \frac{-1}{(1+x^2)^{1/2}} + \frac{1}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{1-3(1+x^2)}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{-3x^2-2}{3(1+x^2)^{3/2}} + C \\ (x,y) = (0,1): &&-1 &= -\frac23 + C \\ \Rightarrow && C &= -\frac13 \\ \Rightarrow && \frac1y &= \frac{3x^2+2}{3(1+x^2)^{3/2}} + \frac13 \end{align*} \begin{align*} y &= \frac{1}{\frac13 +\frac{3x^2+2}{3(1+x^2)^{3/2}} } \\ &= \frac{3}{1+ \frac{3x^2+2}{3(1+x^2)^{3/2}}} \\ &= \frac{3}{1+ \frac{3}{x} + \cdots} \\ &\approx 3 - \frac{9}{x} \end{align*}

TikZ diagram
\begin{align*} && \frac {\d z} {\d x} &= \frac{x^3z^3}{2(1 + x^2)^{5/2}} \\ \Rightarrow && \int \frac{1}{z^3} \d z &= \int \frac{x^3}{2(1+x^2)^{5/2}} \\ && -\frac{1}{2z^2} &= -\frac{3x^2+2}{3(1+x^2)^{3/2}} - C \\ (x,z) = (0, \pm 1): && \frac{1}{2} &= \frac{2}{3} + C \\ \Rightarrow && C &= -\frac16 \\ \Rightarrow && \frac{1}{z^2} &= \frac{6x^2+4}{3(1+x^2)^{3/2}} - \frac13 \end{align*} So as \(x \to \infty\) \(z \sim \pm (3 + \frac{2}{x} + \cdots)\) and so:
TikZ diagram

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2002 Paper 3 Q6
D: 1700.0 B: 1484.9
differential equation first order separation of variables curve sketching implicit solution initial value problem fourth power

Find all the solution curves of the differential equation \[ y^4 \l {\mathrm{d}y \over \mathrm{d}x }\r^{\! \! 4} = \l y^2 - 1 \r^2 \] that pass through either of the points

  1. \(\l 0, \, \frac{1}{2}\sqrt3 \r\),
  2. \(\l 0, \, \frac{1}{2}\sqrt5 \r\).
Show also that \(y = 1\) and \(y = -1\) are solutions of the differential equation. Sketch all these solution curves on a single set of axes.

Solution: \begin{align*} && y^4 \left (\frac{\d y}{\d x} \right)^4 &= (y^2 - 1)^2 \\ \Rightarrow && y^2 \left (\frac{\d y}{\d x} \right)^2 &= |y^2 - 1| \\ && y \left (\frac{\d y}{\d x} \right) &= \pm \sqrt{|y^2-1|} \\ \Rightarrow &&\int \frac{y}{\sqrt{|y^2-1|}} \d y &= \int \pm 1 \d x \\ \Rightarrow && \pm \sqrt{|y^2-1|} &= \pm x + C \\ \end{align*}

  1. Since \(y^2 < 1\), our solution curve should be of the from \(-\sqrt{1-y^2} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{3})\), we obtain \(-\tfrac12 = C\), therefore our solution curves are \(\pm x = \frac12 - \sqrt{1-y^2}\)
  2. Since \(y^2 > 1\), our solution curve should be of the from \(\sqrt{y^2-1} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{5})\), we obtain \(\tfrac12 = C\), therefore our solution curves are \(\pm x = \sqrt{y^2-1}-\frac12\)
Clearly if \(y = \pm 1\), \(y'=0\) and the equation is satisfied.
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2000 Paper 2 Q8
D: 1600.0 B: 1500.1
differential equation first order separable equation exponential function asymptotic behaviour initial value problem integration

  1. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} + 4x\e^{-x^2} {(y+3)}^{\frac12} = 0 \qquad (x \ge 0), \] that satisfies the condition \(y=6\) when \(x=0\). Find \(y\) in terms of \(x\) and show that \(y\to1\) as \(x \to \infty\).
  2. Let \(y\) be any solution of the differential equation \[ \frac{\d y}{\d x} -x\e^{6 x^2} (y+3)^{1-k} = 0 \qquad (x \ge 0). \] %that satisfies the condition \(y=6\) %when \(x=0\). Find a value of \(k\) such that, as \(x \to \infty\), \(\e^{-3x^2}y\) tends to a finite non-zero limit, which you should determine.
\noindent [The approximations, valid for small \(\theta\), \(\sin\theta \approx \theta\) and \(\cos\theta \approx 1-{\textstyle\frac12}\,\theta^2\) may be assumed.]

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1999 Paper 3 Q8
D: 1700.0 B: 1516.0
second order differential equation SHM piecewise function continuity conditions initial value problem infinite series integration trigonometric

The function \(y(x)\) is defined for \(x\ge0\) and satisfies the conditions \[ y=0 \mbox{ \ \ and \ \ } \frac{\d y}{\d x}=1 \mbox{ \ \ at \(x=0\)}. \] When \(x\) is in the range \(2(n-1)\pi< x <2n\pi\), where \(n\) is a positive integer, \(y(t)\) satisfies the differential equation $$ {\d^2y \over \d x^2} + n^2 y=0. $$ Both \(y\) and \(\displaystyle \frac{\d y}{\d x} \) are continuous at \(x=2n\pi\) for \(n=0,\; 1,\;2,\; \ldots\;\).

  1. Find \(y(x)\) for \(0\le x \le 2\pi\).
  2. Show that \(y(x) = \frac12 \sin 2x \) for \(2\pi\le x\le 4\pi\), and find \(y(x)\) for all \(x\ge0\).
  3. Show that $$ \int_0^\infty y^2 \,\d x = \pi \sum_{n=1}^\infty {1\over n^2} \,. $$

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1990 Paper 1 Q7
D: 1500.0 B: 1500.0
differential equation first order integrating factor product substitution initial value problem domain of validity linear ODE

Let \(y,u,v,P\) and \(Q\) all be functions of \(x\). Show that the substitution \(y=uv\) in the differential equation \[ \frac{\mathrm{d}y}{\mathrm{d}x}+Py=Q \] leads to an equation for \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) in terms of \(x,Q\) and \(u\), provided that \(u\) satisfies a suitable first order differential equation. Hence or otherwise solve \[ \frac{\mathrm{d}y}{\mathrm{d}x}-\frac{2y}{x+1}=\left(x+1\right)^{\frac{5}{2}}, \] given that \(y(1)=0\). For what set of values of \(x\) is the solution valid?

Solution: Suppose \(y = uv\) then and suppose \(\frac{\d u}{\d x} + P u = 0\) then \begin{align*} && \frac{\d y}{\d x} + Py &= Q \\ && uv' + u'v + Puv &= Q \\ && uv' &= Q \\ && \frac{\d v}{\d x} &= \frac{Q}{u} \end{align*} Consider \begin{align*} && 0 &= \frac{\d u}{\d x} - \frac{2u}{x+1} \\ \Rightarrow && \ln u &= 2\ln (1 + x) + C \\ \Rightarrow && u &= A(1+x)^2 \end{align*} and \begin{align*} && \frac{\d v}{\d x} &= \frac1{A}(x+1)^{\frac12} \\ \Rightarrow && v &= \frac2{3A}(x+1)^{\frac32} + k \\ \Rightarrow && y &= \frac23(x+1)^\frac72 + k(x+1)^2 \\ && 0 &= y(1) \\ &&&= \frac23 2^{7/2}+k2^2 \\ \Rightarrow && k &= -\frac{2^{5/2}}{3} \\ \Rightarrow && y &= \frac23 (x+1)^{7/2} - \frac{2^{5/2}}{3}(x+1)^2 \end{align*}

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1987 Paper 1 Q3
D: 1500.0 B: 1500.0
differential equation first order integrating factor substitution homogeneous equation asymptotic behaviour initial value problem

By substituting \(y(x)=xv(x)\) in the differential equation \[ x^{3}\frac{\mathrm{d}v}{\mathrm{d}x}+x^{2}v=\frac{1+x^{2}v^{2}}{\left(1+x^{2}\right)v}, \] or otherwise, find the solution \(v(x)\) that satisfies \(v=1\) when \(x=1\). What value does this solution approach when \(x\) becomes large?

Solution: Let \(y = xv\) then \(y' = v + xv'\) and so \(x^2y' = x^2v + x^3v'\) Our differential equation is now: \begin{align*} && x^2 y' &= \frac{1+y^2}{(1+x^2)\frac{y}{x}} \\ \Rightarrow && xy' &= \frac{(1+y^2)}{(1+x^2)y} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x(1+x^2)} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x} - \frac{x}{1+x^2} \\ \Rightarrow && \frac12 \ln(1+y^2) &= \ln x - \frac12 \ln(1+x^2) + C \\ \Rightarrow && \frac12 \ln (1 + y^2) &= \frac12 \ln \l \frac{x^2}{1+x^2}\r + C \\ \Rightarrow && 1+y^2 &= \frac{Dx^2}{1+x^2} \\ \Rightarrow && D &= 4 \quad \quad: (x = 1, v = 1, y = 1) \\ \Rightarrow && 1 + x^2v^2&= \frac{4x^2}{1+x^2}\\ \Rightarrow && v^2 &= \frac{3x^2-1}{x^2(1+x^2)} \\ \Rightarrow && v &= \sqrt{\frac{3x^2-1}{x^2(1+x^2)}} \\ \end{align*} As \(x \to \infty\), \(v \to 0\)

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