Problems
Filters
3 problems found
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).
2011 Paper 3 Q1
- Find the general solution of the differential equation \[ \frac{\d u}{\d x} - \left(\frac { x +2}{x+1}\right)u =0\,. \]
- Show that substituting\(y=z\e^{-x}\) (where \(z\) is a function of \(x\)) into the second order differential equation \[ (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y = 0 \tag{\(*\)} \] leads to a first order differential equation for \(\dfrac{\d z}{\d x}\,\). Find \(z\) and hence show that the general solution of \((*)\) is \[ y= Ax + B\e^{-x}\,, \] where \(A\) and \(B\) are arbitrary constants.
- Find the general solution of the differential equation \[ (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y = (x+1)^2 . \]
Solution:
- \begin{align*} && 0 &= \frac{\d u}{\d x} - \left ( \frac{x+2}{x+1} \right)u \\ \Rightarrow && \int \frac1u \d u &= \int 1 + \frac1{x+1} \d x \\ \Rightarrow && \ln |u| &= x + \ln |x+1| + C \\ \Rightarrow && u &= A(x+1)e^x \end{align*}
- If \(y = ze^{-x}\), \(y' = (z'-z)e^{-x}\), \(y'' = (z''-2z'+z)e^{-x}\) \begin{align*} && 0 &= (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y \\ y = ze^{-x}: && 0 &= (x+1) \left ( \frac{\d^2 z}{\d x^2} - 2\frac{\d z}{\d x} +z\right)e^{-x} +x \left ( \frac{\d z}{\d x} -z\right)e^{-x} - ze^{-x} \\ &&&= (x+1) \frac{\d^2 z}{\d x^2} -(x+2)\frac{\d z}{\d x} \\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{\d z}{\d x}\right) &= \left ( \frac{x+2}{x+1}\right) \frac{\d z}{\d x} \end{align*} Therefore \(\frac{\d z}{\d x} = A(x+1)e^x \) and so \begin{align*} z &= A \int (x+1)e^{x} \d x \\ &= A \left ( \left [ (x+1)e^x\right] - \int e^x \d x \right) \\ &= A(x+1)e^x - Ae^x + B \\ y &= Ax + Be^{-x} \end{align*}
- We have found the complementary solution. To find a particular integral consider \(y = ax^2 + bx + c\), then \(y' = 2ax+b, y'' = 2a\) and we have \begin{align*} && x^2+2x+1 &= 2a(x+1) + x(2ax+b) - (ax^2+bx+c) \\ \Rightarrow && x^2+2x+1 &= ax^2+ 2ax + 2a-c \\ \Rightarrow && a = 1, &c=1 \end{align*} so the general solution should be \[ y = Ax + Be^{-x} + x^2+1 \]
1990 Paper 1 Q7
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*}