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2003 Paper 3 Q8
D: 1700.0 B: 1516.0
first order differential equation implicit differentiation exact equation integrating factor homogeneous ODE curve gradient

  1. Show that the gradient at a point \(\l x\,, \, y \r\) on the curve \[ \l y + 2x \r^3 \l y - 4x \r = c\;, \] where \(c\) is a constant, is given by \[ \frac{\d y}{\d x} = \frac{16 x -y}{2y-5x} \;. \]
  2. By considering the derivative with respect to \(x\) of \(\l y + ax \r^n \l y + bx \r\,\), or otherwise, find the general solution of the differential equation \[ \frac{\mathrm{d}y}{ \mathrm{d}x} = \frac{10x - 4y}{ 3x - y}\;. \]

Solution:

  1. \begin{align*} && c &= \left (y+2x \right)^3\left (y-4x \right) \\ \Rightarrow && 0 &= 3\left (y+2x \right)^2\left (y-4x \right)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right)^3 \left ( \frac{\d y}{\d x} - 4 \right) \\ \Rightarrow && 0 &= 3(y-4x)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right) \left ( \frac{\d y}{\d x} - 4 \right) \\ \Rightarrow &&&= \frac{\d y}{\d x} \left (3(y-4x) + (y+2x) \right) + 6(y-4x)-4(y+2x) \\ &&&= \frac{\d y}{\d x} \left ( 4y-10x\right) + 2y-32x \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{16x-y}{2y-5x} \end{align*}
  2. \begin{align*} && c &= \left ( y + ax \right)^n \left ( y + bx \right) \\ \Rightarrow && 0 &= n\left ( y + ax \right)^{n-1} \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right)^{n} \left ( \frac{\d y}{ \d x}+b \right) \\ \Rightarrow && 0 &= n \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right) \left ( \frac{\d y}{ \d x}+b \right) \\ &&&= \frac{\d y}{\d x} \left ( (n+1)y + (nb+a)x \right) + an(y+bx) + by+bax \\ &&&= \frac{\d y}{\d x} \left ( (n+1)y + (nb+a)x \right) + (an+b)y+ab(n+1)x \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{(an+b)y+ab(n+1)x}{(n+1)y+(nb+a)x} \end{align*} We must have \(ab = 10, a+b = -7\) so say \(a=-5,b=-2,n=2\) and we have \((y-5x)^2(y-2) = c\) is our general solution to the differential equation

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