Pearson Edexcel IAL Mathematics FP2.5 Second order differential equations Question Bank

Practise solving FP2 second order differential equations through auxiliary equations, particular integrals and variable transformations.

Exam points

solve the auxiliary equation and combine the complementary function with a suitable PI
convert the solution back to x and apply any given y and dy/dx conditions

[Maximum number: 7]

(a)

Hence show that the transformation $t=\ln x$ , where x>0, transforms the differential equation

x^2\frac{d^2y}{dx^2}-2y=1+4\ln x-2(\ln x)^2

into the differential equation

\frac{d^2y}{dt^2}-\frac{dy}{dt}-2y=1+4t-2t^2

[ 1 ]

x^2\frac{d^2y}{dx^2}-2y=e^{2t}e^{-2t}\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)-2y

= =\frac{d^2y}{dt^2}-\frac{dy}{dt}-2y.

Also $\ln x=t$ , so

\frac{d^2y}{dt^2}-\frac{dy}{dt}-2y=1+4t-2t^2

(b)

Solve differential equation (II) to determine y in terms of t.

[ 5 ]

For the complementary function,

m^2-m-2=0

so m=2,-1 and

y=Ae^{2t}+Be^{-t}

For a particular integral, let $y=at^2+bt+c$ . Then

\frac{dy}{dt}=2at+b,\qquad \frac{d^2y}{dt^2}=2a

Substitution gives

2a-2at-b-2at^2-2bt-2c=1+4t-2t^2

Comparing coefficients:

a=1,\qquad b=-3,\qquad c=2

Therefore

y=Ae^{2t}+Be^{-t}+t^2-3t+2

(c)

Hence determine the general solution of differential equation (I).

[ 1 ]

Since $t=\ln x$ ,

e^{2t}=x^2,\qquad e^{-t}=x^{-1}

Therefore the general solution is

y=Ax^2+Bx^{-1}+(\ln x)^2-3\ln x+2

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