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Second order equations reducible by substitutionEdexcel International A Level Further Maths: Mind map

Method
$x=\mathrm{e}^t$
$y=zf(x)$

Reduction by substitution

to constant coefficients

x=etx=\mathrm{e}^ty=zf(x)y=zf(x)Euler
Solving
Conditions
Exam tips

Exam questions on Second order equations reducible by substitution

  1. A differential equation for yy in terms of xx (x>0x>0) is to be transformed using the substitution x=etx=\mathrm{e}^t. A dot denotes differentiation with respect to tt, so y˙=dydt\dot y=\frac{dy}{dt} and y¨=d2ydt2\ddot y=\frac{d^2y}{dt^2}.
    The equation x2d2ydx2−2xdydx+2y=0x^2\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=0 transforms into y¨−3y˙+2y=0\ddot y-3\dot y+2y=0. Find the general solution for yy in terms of xx.2 marks
  2. The differential equation d2ydx2+4xdydx+(4x2+2)y=8e−x2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}+(4x^2+2)y=8\mathrm{e}^{-x^2} is to be transformed using the substitution y=z e−x2y=z\,\mathrm{e}^{-x^2}, where zz is a function of xx.
    The equation transforms into d2zdx2=8\frac{d^2z}{dx^2}=8. Find the general solution for yy in terms of xx.2 marks
  3. The differential equation x2d2ydx2+5xdydx+4y=0x^2\frac{d^2y}{dx^2}+5x\frac{dy}{dx}+4y=0, x>0x>0, is to be solved using the substitution x=etx=\mathrm{e}^t.
    Show that the substitution transforms the equation into d2ydt2+4dydt+4y=0\frac{d^2y}{dt^2}+4\frac{dy}{dt}+4y=0.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).