MT Chapter 10.pdf

LEE KIAN KEONG STPM MATHEMATICS (T) 10: Differential Equations

10 Differential Equations

1. [STPM ]Variables t and y, with t 0 and y > 0, are related by

dy

dt=

1

2y(2 y),

with the condition y = 1 when t = 0. Show that y =2et

1 + et. [8 marks]

Show that y 2 when t. Sketch the graph of y versus t. [4 marks]Find the difference in the values of t when y changes from

6

5to

9

5. [3 marks]

[Answer : 1.7917]

2. [STPM ]

A curve passes through point (2, 0) such that its gradientdy

dxat point (x, y) satisfies equation (x2 3)dy

dx=

4x(3 + y). Show that the equation of the curve is y = 3(x2 2)(x2 4). [6 marks]Sketch the curve. [3 marks]

Find the area of the region bounded by the curve and the x-axis. [6 marks]

[Answer :48

5(6

2 4)]

3. [STPM ]

(a) Variables t and v, with 0 < t < 2, is related by the differential equation

tdv

dt= v2 v,

with the condition v = 2 when t = 1. Find v in terms of t and sketch the graph of v versus t. [8 marks]

(b) Show that

(1 + x2)32d

dx

(y

1 + x2

)= (1 + x2)

dy

dx xy,

where y is a function of x. Hence, solve the differential equation

(1 + x2)dy

dx xy = x(1 + x2),

with the condition y = 1 when x = 0. [7 marks]

[Answer : (a) v =2

2 t ; (b) y = 1 + x2]

4. [STPM ]

(a) Solve the differential equation

(1 x)(1 + x2)dydx

+ (2 x+ x2)y = (1 x)2, x < 1,

with the condition y = 3 when x = 0. [8 marks]

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5. [STPM ]Solve the differential equation

xdy

dx= (x+ 1)(y + 1),

with x > 0 and y > 1, and y = 2 when x = 1. Give your answer in the form y = f(x).

[Answer : y = 3xex1 1]

6. [STPM ]Find the particular solution for the differential equation

dy

dx+

x 2x(x 1)y =

1

x2(x 1) .

that satisfies the boundary condition y =3

4when x = 2. [8 marks]

[Answer : y =2x 1x2

]

7. [STPM ]The variables t and x are connected by

dx

dt= 2t(x 1),

where x 6= 1. Find x in terms of t if x = 2 when t = 1. [5 marks]

[Answer : x = et21 + 1]

8. [STPM ]

Using the substitution y =v

x2, show that the differential equation

dy

dx+ y2 = 2y

x

may be reduced todv

dx= v

2

x2.

[3 marks]

Hence, find the general solution of the original differential equation. [4 marks]

[Answer : y =1

Ax2 x ]

9. [STPM ]

Show thatd

dx(ln tanx) =

2

sin 2x. [2 marks]

Hence, find the solution of the differential equation

(sin 2x)dy

dx= 2y(1 y)

for which y =1

3when x =

1

4pi. Express y explicitly in terms of x in your answer. [8 marks]

[Answer : y =tanx

2 + tanx]

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10. [STPM ]Find the general solution of the differential equation

xdy

dx 3y = x3.

[4 marks]

Find the particular solution given that y has a minimum value when x = 1. [3 marks]

Sketch the graph of this particular solution. [3 marks]

[Answer : y = x3 lnx + Cx3 ; y = x3 lnx 13x3]


xdy

dx= y2 y 2.

[6 marks]

[Answer : y =2 + Ax3

1Ax3 ]

12. [STPM ]The variables x and y, where x > 0, satisfy the differential equation

x2dy

dx= y2 xy.

Using the substitution y = ux, show that the given differential equation may be reduced to

xdu

dx= u2 2u.

Hence, show that the general solution of the given differential equation may be expressed in the form y =2x

1 +Ax2,

where A is an arbitrary constant. [10 marks]

Find the equation of the solution curve which passes through the point (1,4) and sketch this solution curve.[4 marks]

[Answer : y =4x

2 x2 ]

13. [STPM ]Show that the substitution u = x2 + y transforms the differential equation

(1 x)dydx

+ 2y + 2x = 0

into the differential equation

(1 x)dudx

= 2u.[3 marks]

14. [STPM ]Find the particular solution of the differential equation

exdy

dx y2(x+ 1) = 0

for which y = 1 when x = 0. Hence, express y in terms of x. [7 marks]

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[Answer : y =ex

2 + x ex ]

15. [STPM ]

Using the substitution z =1

y, show that the differential equation

dy

dx 2y

x= y2

may be reduced todz

dx+

2z

x= 1.

[2 marks]

Hence, find the particular solution y in terms of x for the differential equation given that y = 3 when x = 1.[6 marks]

Sketch the graph y. [3 marks]

[Answer : y =3x2

2 x3 ]


1

x

dy

dx=

2 lnx

cos y.

[5 marks]

[Answer : sin y = x2 lnx 12x2 + c]

17. [STPM ]Using the substitution y = vx, show that the differential equation

xydy

dx x2 y2 = 0

may be reduced to

vxdv

dx= 1.

[3 marks]

Hence, find the particular solution that satisfies y = 2 and x = 1. [6 marks]

[Answer : y2 = 2x2(lnx+ 2)]

18. [STPM ]The variables x and y, where x > 0 and y > 0, satisfy the differential equation

dy

dx=y(y + x)

x(y x) .

Show that the substitution y = ux transforms the given differential equation into the differential equation

du

dx=

2u

x(u 1) .

[3 marks]

Hence, find the solution of the given differential equation for which y = 2 when x =1

2. [6 marks]

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[Answer : y = 4x+ x lnxy or xy = eyx4]

19. [STPM ]Differentiate yex with respect to x. Hence, find the solution of the differential equation

dy

dx y = ex cosx

for which y = 1 when x = 0. [6 marks]

[Answer : exdy

dx; yex = sinx + 1]

20. [STPM ]The variables x and y, where x > 0, satisfy the differential equation

x2dy

dx= 2xy + y2.

Using the substitution y = ux, show that the given differential equation can be transformed into

xdu

dx= u+ u2.

[3 marks]

Show that the general solution of the transformed differential equation can be expressed as u =x

A x , where Ais an arbitrary constant. [7 marks]

Hence, find the particular solution of the given differential equation which satisfies the condition that y = 2 whenx = 1. [3 marks]

[Answer : y =2x2

3 2x ]

21. [STPM ]Find the solution of the differential equation

dy

dx 2xy

1 + x2= ex(1 + x2)

given that y = 3 when x = 0. [8 marks]

[Answer : y = (ex + 2)(1 + x2)]

22. [STPM ]

Show that e

x2x(x1)dx =

x2

x 1 . [4 marks]Hence, find the particular solution of the differential equation

dy

dx+

x 2x(x 1)y =

1

x2(x 1)

which satisfies the boundary condition y =3

4when x = 2. [4 marks]

[Answer : y =2x 1x2

]

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23. [STPM ]The variables x and y, where x, y > 0, are related by the differential equation

dy

dx+ y2 = 2y

x.

Using the substitution y =u

x2, show that the differential equation may be reduced to

du

dx= u

2

x2.

[3 marks]

Solve this differential equation, and hence, find y, in terms of x, with the condition that y = 1 when x = 1.[6 marks]

[Answer : y =1

x(2x 1) ]

24. [STPM ]Show that the substitution y = ux transform the differential equation

xdy

dx= y 2x cot

(yx

)into the differential equation

xdu

dx= 2 cotu.

[3 marks]

Hence, find the solution of the given differential equation satisfying the condition y = 0 when x = 1. Give youranswer in the form y = f(x). [5 marks]

[Answer : y = x cos1 x2]

25. [STPM ]Using the substitution u = ln y, show that the non-linear differential equation

xdy

dx+ (3x+ 1)y ln y = ye2x

can be transformed into the linear differential equation

xdu

dx+ (3x+ 1)u = e2x.

[4 marks]

Solve this linear differential equation, and hence, find the solution of the original non-linear differential equation,given that y = 1 when x = 1. [9 marks]

Find the limiting value of y as x. [2 marks]

[Answer : y = ee2xx e13x

x ; 1]

26. [STPM ]The variables x and y, where x, y > 0 are related by the differential equation

xydy

dx+ y2 = 3x4.

Show that the substitution u =y

x2transforms the above differential equation into

xdu

dx= 3

(1 u2u

),

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and find u2 in terms of x. [9 marks]

Hence, find the particular solution of the original differential equation which satisfies the condition y = 2 whenx = 1. [3 marks]

[Answer : u2 = 1 Ax6

; y2 = x4 +3

x2]

27. [STPM ]Solve the differential equation

dy

dx 2xy = x2e2x.

[6 marks]

[Answer : y =1

2x2e2x + Cx2]

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1 Functions2 Sequences and Series3 Matrices4 Complex Numbers5 Analytic Geometry6 Vectors7 Limits and Continuity8 Differentiation9 Integration10 Differential Equations11 Maclaurin Series12 Numerical Methods13 Data Description14 Probability15 Probability Distributions16 Sampling and Estimation17 Hypothesis Testing18 Chi-squared Tests

MT Chapter 10.pdf

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