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2019 Paper 2 Q6
Note: You may assume that if the functions \(y_1(x)\) and \(y_2(x)\) both satisfy one of the differential equations in this question, then the curves \(y = y_1(x)\) and \(y = y_2(x)\) do not intersect.
- Find the solution of the differential equation $$\frac{dy}{dx} = y + x + 1$$ that has the form \(y = mx + c\), where \(m\) and \(c\) are constants. Let \(y_3(x)\) be the solution of this differential equation with \(y_3(0) = k\). Show that any stationary point on the curve \(y = y_3(x)\) lies on the line \(y = -x - 1\). Deduce that solution curves with \(k < -2\) cannot have any stationary points. Show further that any stationary point on the solution curve is a local minimum. Use the substitution \(Y = y + x\) to solve the differential equation, and sketch, on the same axes, the solutions with \(k = 0\), \(k = -2\) and \(k = -3\).
- Find the two solutions of the differential equation $$\frac{dy}{dx} = x^2 + y^2 - 2xy - 4x + 4y + 3$$ that have the form \(y = mx + c\). Let \(y_4(x)\) be the solution of this differential equation with \(y_4(0) = -2\). (Do not attempt to find this solution.) Show that any stationary point on the curve \(y = y_4(x)\) lies on one of two lines that you should identify. What can be said about the gradient of the curve at points between these lines? Sketch the curve \(y = y_4(x)\). You should include on your sketch the two straight line solutions and the two lines of stationary points.
Solution:
- Looking for solution of the form \(y = mx+c\) we find that \(m = mx+c+x+1 \Rightarrow m = -1, c = -2\).
At stationary points \(\frac{\d y}{\d x} = 0 \Rightarrow y = -x-1\).
If \(y_3(0)= k < -2\) then the solution curve lies below \(y = -x-2\) and therefore it cannot cross \(y = -x -2\) to reach \(y = -x-1\) for a stationary point.
Suppose \(Y = y+x\) then \(\frac{\d Y}{\d x} = \frac{\d y}{\d x} + 1=Y+2 \Rightarrow Y = Ae^x-2 \Rightarrow y= (k+2)e^x-x-2\)
- \(\,\)
\begin{align*}
&& m &= x^2 + (mx+c)^2 -2x(mx+c) - 4x+4(mx+c) + 3 \\
&&0&= (m^2-2m+1)x^2+(2mc-2c-4+4m)x + (c^2+4c+3-m)\\
\Rightarrow && m &= 1 \\
\Rightarrow && 0 &= c^2+4c+2 \\
\Rightarrow &&&= (c+2)^2-2 \\
\Rightarrow && c &= -2 \pm \sqrt{2}
\end{align*}
Therefore the lines are \(y = x -2-\sqrt{2}\) and \(y = x -2+\sqrt{2}\).
Any stationary point will satisfy \(y' = 0\), ie \(0 = x^2+y^2-2xy-4x+4y+3 = (x-y)^2-4(x-y)+3 = (x-y-3)(x-y-1)\) therefore they must lie on \(y = x-1\) or \(y = x-3\). Any point between these lines must have negative gradient (since one factor is positive and one factor is negative).
2007 Paper 3 Q8
- Find functions \({\rm a}(x)\) and \({\rm b}(x)\) such that \(u=x\) and \(u=\e^{-x}\) both satisfy the equation $$\dfrac{\d^2u}{\d x^2} +{\rm a}(x) \dfrac{\d u}{\d x} + {\rm b} (x)u=0\,.$$ For these functions \({\rm a}(x)\) and \({\rm b}(x)\), write down the general solution of the equation. Show that the substitution \(y= \dfrac 1 {3u} \dfrac {\d u}{\d x}\) transforms the equation \[ \frac{\d y}{\d x} +3y^2 + \frac {x} {1+x} y = \frac 1 {3(1+x)} \tag{\(*\)} \] into \[ \frac{\d^2 u}{\d x^2} +\frac x{1+x} \frac{\d u}{\d x} - \frac 1 {1+x} u=0 \] and hence show that the solution of equation (\(*\)) that satisfies \(y=0\) at \(x=0\) is given by \(y = \dfrac{1-\e^{-x}}{3(x+\e^{-x})}\).
- Find the solution of the equation $$ \frac{\d y}{\d x} +y^2 + \frac x {1-x} y = \frac 1 {1-x} $$ that satisfies \(y=2\) at \(x=0\).
2004 Paper 3 Q8
Show that if \[ {\mathrm{d}y \over \mathrm{d} x}=\f(x)y + {\g(x) \over y} \] then the substitution \(u = y^2\) gives a linear differential equation for \(u(x)\,\). Hence or otherwise solve the differential equation \[ {\mathrm{d}y \over \mathrm{d} x}={y \over x} - {1 \over y}\;. \] Determine the solution curves of this equation which pass through \((1 \,, 1)\,\), \((2\, , 2)\) and \((4 \, , 4)\) and sketch graphs of all three curves on the same axes.
Solution: \begin{align*} && \frac{\d y}{\d x} &= f(x) y + \frac{g(x)}{y} \\ && y \frac{\d y}{\d x} &= f(x) y^2 + g(x) \\ u = y^2: && \frac12 \frac{\d u}{\d x} &= f(x) u + g(x) \end{align*} Which is a linear differential equation for \(u\). \begin{align*} && \frac12 u' &= \frac1x u -1 \\ \Rightarrow && u' - \frac2xu &= -1 \\ \Rightarrow && \frac{1}{x^2} u' - \frac{2}{x^3} u &= -\frac{1}{x^2} \\ \Rightarrow && (\frac{u}{x^2})' &= - \frac{1}{x^2} \\ \Rightarrow && \frac{u}{x^2} &= \frac1x + C \\ \Rightarrow && u &= Cx^2 + x \\ \Rightarrow && y^2 &= Cx^2 + x \end{align*} If \((1,1)\) is on the curve then \(1 = C + 1 \Rightarrow C = 0 \Rightarrow y^2 = x\). If \((2,2)\) is on the curve then \(4 = 4C + 2 \Rightarrow C = \frac12 \Rightarrow y^2 = x + \frac12 x^2\). If \((3,3)\) is on the curve then \(9 = 9C + 3 \Rightarrow C = \frac23 \Rightarrow y^2 = x + \frac23 x^2\)
1998 Paper 3 Q3
The value \(V_N\) of a bond after \(N\) days is determined by the equation $$ V_{N+1} = (1+c) V_{N} -d \qquad (c>0, \ d>0), $$ where \(c\) and \(d\) are given constants. By looking for solutions of the form \(V_T= A k^T + B\) for some constants \(A,B\) and \(k\), or otherwise, find \(V_N\) in terms of \(V_0\). What is the solution for \(c=0\)? Show that this is the limit (for fixed \(N\)) as \(c\rightarrow 0\) of your solution for \(c>0\).
Solution: Suppose \(V_T = Ak^T + B\), then \begin{align*} && Ak^{T+1}+B &= (1+c)(Ak^T+B) - d \\ \Rightarrow && (k-1-c)Ak^T &= cB -d \\ \Rightarrow && k &= 1+c \\ && B &= \frac{d}{c} \\ && A &= V_0 - B \\ \Rightarrow && V_N &= (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \end{align*} When \(c = 0\), \(V_{N+1} = V_N - d \Rightarrow V_N = V_0 - Nd\). \begin{align*} \lim_{c \to 0} \left ( (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \right) &= \lim_{c \to 0} \left ( \frac{(V_0 c-d)(1+c)^N + d}{c} \right ) \\ &= \lim_{c \to 0} \left ( \frac{V_0c - d-Ncd+NV_0c^2 + o(c^2) + d}{c} \right ) \\ &= \lim_{c \to 0} \left ( V_0 - Nd + o(c) \right ) \\ &= V_0 - Nd \end{align*}
1990 Paper 3 Q8
Let \(P,Q\) and \(R\) be functions of \(x\). Prove that, for any function \(y\) of \(x\), the function \[ Py''+Qy'+Ry \] can be written in the form \(\dfrac{\mathrm{d}}{\mathrm{d}x}(py'+qy),\) where \(p\) and \(q\) are functions of \(x\), if and only if \(P''-Q'+R=0.\) Solve the differential equation \[ (x-x^{4})y''+(1-7x^{3})y'-9x^{2}y=(x^{3}+3x^2)\mathrm{e}^{x}, \] given that when \(x=2,y=2\mathrm{e}^{2}\) and \(y'=0.\)
Solution: Suppose \(Py'' + Qy' + Ry = \frac{\d}{\d x}(p y' + qy)\), then \begin{align*} Py'' + Qy' + Ry &= \frac{\d}{\d x}(p y' + qy) \\ &= py'' + p'y' + qy' + q' y \\ &= py'' + (p'+q)y' + q' y \end{align*} Therefore \(P = p, Q = p'+q, R = q'\), Therefore \(q = Q-P'\) and \(R = Q'-P''\) or \(P'' -Q'+R = 0\). \((\Rightarrow)\) Suppose it can be written in that form, then the logic we have applied shows that equation is true. \((\Leftarrow)\) Suppose \(P''-Q'+R = 0\), then letting \(p = P, q = Q-P'\) we find functions of the form which will be expressed correctly. Notice that if \(P = x-x^4, Q = (1-7x^3), R = -9x^2\) then: \begin{align*} P'' - Q' + R &= -12x^2+21x^2-9x^2 \\ &= 0 \end{align*} Therefore we can write our second order ODE as: \begin{align*} && (x^{3}+3x)\mathrm{e}^{x} &= \frac{\d}{\d x} \left ( (x-x^4) y' +(1-7x^3-(1-4x^3))y \right) \\ &&&= \frac{\d}{\d x} \left ( (x-x^4) y' -3x^3y \right) \end{align*} Suppose \(z = (x-x^4)y' - 3x^2y\), then \(z = (2-2^4) \cdot 0 - 3 \cdot 2^2 \cdot 2e^2 = -24e^2\) when \(x = 2\). and we have: \begin{align*} && \frac{\d z}{\d x} &= (x^3+3x^2)e^x \\ \Rightarrow && z &= \int (x^3+3x^2)e^x \d x \\ &&&= x^3 e^x+c \\ \Rightarrow && -48e^2 &= e^2(8) + c \\ \Rightarrow && c &= -56e^2 \\ \Rightarrow && z &= e^x(x^3)-56e^2 \\ \end{align*} So our differential equation is: \begin{align*} && (x-x^4)y'-3x^3 y &= x^3e^x -5 6e^2 \\ \Rightarrow && (1-x^3)y' -3x^2y &= x^2 e^x - \frac{6e^2}{x} \\ \Rightarrow && \frac{\d }{\d x} \left ( (1-x^3)y \right) &= x^2e^x - \frac{56e^2}{x} \\ \Rightarrow && (1-x^3)y &= (x^2-2x+2)e^x - 56e^2 \ln x + k \\ \underbrace{\Rightarrow}_{x=2} && (1-2^3)2e^2 &= (2^2-2\cdot2 + 2)e^2 -56e^2 \ln 2 + k \\ \Rightarrow && k &= -16e^2+56 \ln 2 \cdot e^2 \\ \Rightarrow && y &= \frac{(x^2-2x+2)e^x - 56e^2 \ln x -16e^2+56 \ln 2 \cdot e^2}{(1-x^3)} \end{align*}