Pearson Edexcel IAL Mathematics FP2.4 First order differential equations Question Bank

Practise solving FP2 first order differential equations using integrating factors, substitutions and initial conditions to reach general or particular solutions.

Exam points

choose and apply an integrating factor before writing z, y or y^2 in the required form
use a given substitution to transform dy/dx equations into a solvable linear equation
apply an initial condition to fix the constant and state the particular solution as y=f(x)

Covered syllabus points

FP2.4.1 - Further solution of first order FP2.4.2 - First order linear differential FP2.4.3 - Reducible first order differential equations

[Maximum number: 10]

(a)

Show that the transformation v=y-2 x transforms the differential equation

\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y x(y-4 x)=2-8 x^{3}

into the differential equation

\frac{\mathrm{d} v}{\mathrm{~d} x}=-2 x v^{2}

[ 4 ]

$\frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} x}-2$

B1

$\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y x(y-4 x)=2-8 x^{3} \rightarrow \frac{\mathrm{~d} v}{\mathrm{~d} x}+2+2(v+2 x) x(v+2 x-4 x)=2-8 x^{3}$ }

$\rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}+2+2 x\left(v^{2}-4 x^{2}\right)=2-8 x^{3}$

М1

$\rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}=-2 x v^{2} *$

A1*

(4)

(b)

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

[ 1 ]

$y=2 x+\frac{1}{x^{2}+c}$

B1ft

(1)

(c)

Sketch the solution curve that passes through the point (-1,-1).

On your sketch show clearly the equation of any horizontal or vertical asymptotes.

You do not need to find the coordinates of any intercepts with the coordinate axes or the coordinates of any stationary points.

Table for Question (c) — Edexcel A-Level Mathematics A2

[ 5 ]

-1=2(-1)+\frac{1}{1+c}\Rightarrow c=0

y=2x+\frac{1}{x^2}

Sketch attempted for the correct curve.

Figure for Answer to Question (c) — Edexcel A-Level Mathematics A2

Two branches, both tending to infinity as x tends to ±infinity, with a minimum in the first quadrant.

B1ft

x=0\text{ is the vertical asymptote}

[Maximum number: 9]

(a)

Show that the substitution $y^{2}=\frac{1}{t}$ transforms the differential equation

\frac{\mathrm{d} y}{\mathrm{~d} x}+y=x y^{3}

into the differential equation

\frac{\mathrm{d} t}{\mathrm{~d} x}-2 t=-2 x

[ 3 ]

Since $y^2=\frac1t$ , $y=t^{-1/2}$ , so

\frac{dy}{dx}=-\frac12t^{-3/2}\frac{dt}{dx}

Substituting into $\frac{dy}{dx}+y=xy^3$ :

-\frac12t^{-3/2}\frac{dt}{dx}+t^{-1/2}=xt^{-3/2}

Multiplying by $-2t^{3/2}$ gives

\frac{dt}{dx}-2t=-2x

(b)

Solve differential equation (II) and determine $y^{2}$ in terms of x.

[ 6 ]

The integrating factor is

I=e^{-\int 2\,dx}=e^{-2x}

te^{-2x}=-2\int xe^{-2x}\,dx

= =xe^{-2x}-\int e^{-2x}\,dx

= =xe^{-2x}+\frac12e^{-2x}+c.

Hence

t=x+\frac12+ce^{2x}

Since $t=\frac1{y^2}$ ,

y^2=\frac{1}{x+\frac12+ce^{2x}}

All question bank results loaded

Covered syllabus points

FP2.4 - First order differential equations question 1

Question (a)

Question (b)

Question (c)

FP2.4 - First order differential equations question 2

Question (a)

Question (b)