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A- LEVEL – MATHEMATICS P 3 Differential Equations
EXERCISE 1 (Answers on page 9-10) (With References) Q1. The variables x and θ satisfy the differential equation : = (x + 2 ) sin^2 2 θ and it is given that x = 0 when θ = 0. Solve the differential equation and calculate the value of x when θ = giving your answer correct to 3 significant figures. [2017/ SP -3/Q8] [W-15 /31/32/Q8]
Q2. The variables x and y satisfy the differential equation : = x ex+y^ and it is given that y = 0 when x= i) Solve the differential equation and obtain an expression for y in terms of x. ii) Explain briefly why x can only take values less than 1.
[M-16 / 32 /Q7]
Q3. The variables x and y satisfy the differential equation: x = y ( 1- 2 x^2 ) and it is given that y = 2 when x = 1. Solve the differential equation and obtain an expression for y in terms of x in a form not involving logarithms.
[S-16 / 31/Q4] Q4. The variables x and θ satisfy the differential equation: ( 3 + cos 2 θ) = x. sin2θ and it is given that x = 3 when θ= i) Solve the differential equation and obtain an expression for x in terms of θ. ii) State the least value taken by x.
[S-16/32/Q6]
9 7 1 6 7 1
Q5. The variables x and y satisfy the differential equation: = e-2y^. tan 2 x , for 0 ≤ x < π and it is given that y = 0 when x = 0. Solve the differential equation and calculate the value of y when x = π
[S-16/33/Q5] Q6. A large field of area 4 km^2 is becoming infected with a soil disease. At time t years the area infected is x km^2 and the rate of growth of the infected area is given by the differential equation: = kx ( 4 – x ), where k is a positive constant. It is given that when t = 0 , x = 0.4 and that when t = 2 , x = 2. i) Solve the differential equation and show that k = ln3. ii) Find the value of t when 90% of the field is infected.
[W-16/31/Q10] Q7. The diagram shows a variable point P with coordinate ( x , y) and the point N which is the foot of the perpendicular from P to x – axis. P moves on a curve such that , For all x ≥ 0 , the gradient of the curve is equal in value to the area of the triangle 0PN, where O is origin. i) State a differential equation satisfied by x and y. The point with coordinates ( 0, 2 ) lies on the curve. ii) Solve the differential equation to obtain the equation of the curve , expressing y in terms of x. Y P (x,y) iii) Sketch the curve. X O N
[W-16/33/Q5]
8 9 2 1 5 1
Q11. Naturalists are managing a wildlife reserve to increase the number of plants of a rare species. The number of plants at time t years is denoted by N , where N is treated as continuous variable. i) It is given that the rate of increase of N with respect to t is proportional to (N – 150 ). Write down a differential equation relating N , t and a constant of proportionality. ii) Initially when t= 0 , the number of plants was 650. It is noted that at a time when there were 900 plants , the number of plants was increasing at a rate of 60 per year. Express N in terms of t. iii) The naturalist had a target of increasing the number of plants from 650 to 2000 within 15 years. Will this target be met? [W-15/33/Q10]
Q12. The variables x and y are related by the differential equation:
=
Given that y = 36 , when x = 0 , find an expression for y in terms of x. [S-14/31/Q4]
Q13. The population of a country at time t years is N millions. At any time, N is assumed to increase at a rate proportional to the product of N and (1 – 0.01N)
when t= 0 , N = 20 and = 0.
i) Treating N and t as continuous variables, show that they satisfy the differential equation. = 0.02 N ( 1 – 0.01N ) ii) Solve the differential equation obtaining an expression for t in terms of N.
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7
6
2
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iii) Find the time at which the population will be double its value at t = 0 [S-14/32/Q9] Q14. The variables x and θ satisfy the differential equation :
2cos^2 θ. = √(2x + 1)
And x = 0 when θ= π. Solve the differential equation and obtain an expression
for x in terms of θ. [S-14/33/Q5] Q15. In a certain country government charges tax on each litre of patrol sold to motorist. The revenue per year is R million dollars when the rate of tax is x dollars per litre. The variation of R with x is modelled by the differential equation:
= R ( ), where R and x are taken to be continuous
variables when x= 0.5 , R = 16. i) Solve the differential equation and obtain an expression for R in terms of x. ii) This model predicts that R cannot exceed a certain amount. Find this maximum value of R. [W-14/31/ 32/Q7] Q16. The variable x and y are related by differential equation:
. sin ( ) i) Find the general solution giving y in terms of x. ii) Given that y = 100 when x = 0, find the value of y when x = 25. [W-14 /33/Q8]
1 7 6 3 6 3
Q19. The variable x and t satisfy the differential equation:
t. = for t > 0 , where k is a constant. When t = 1 , x= 1 and when t = 4 , x = 2 i) Solve the differential equation finding the value of k and obtaining an expression for x in terms of t. ii) State what happens to the value as t becomes large. [S-13/33/Q8] Q20. A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi vertical angle is 60^0 as shown in the diagram. At t = 0, the tank is full and the depth of water is H. At this instant a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to √h , where h is the depth of water at time t. The tank becomes empty when t = 60.
i) Show that h and t satisfy a differential equation of the form: = - A where A is a positive constant. ii) Solve the differential equation given in part (i) and obtain an expression for t in terms of h and H. iii) Find the time at which the depth reaches H. [W-13/31/32/Q10] Q21. A particular solution of the differential equation:
3y^2 = 4 ( y^3 + 1) cos 2 x
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4
6 1
is such that y = 2 when x= 0. The diagram shows a sketch of the graph of this solution of 0 ≤ x ≤ 2 π ; the graph has stationary points at A and B. Find the y- coordinates of A and B , giving each coordinates to 1 decimal plane.
[W-13/33/Q10]
Q22. The variables x and y are related by the differential equation:
= It is given that y = 2 when x=0.
Solve the differential equation and hence find the value of y when x = 0. giving yours answer correct to two decimal places. [S-12/31/Q7] Q23. The variables x and y satisfy the differential equation:
= e2x+y^ and y =0 when x=0.
Solve the differential equation , obtain an expression for y in terms of x. [S-12/32/Q5] Q24. In a certain chemical process a substance A reacts with another substance B. The masses in grams of A and B present at time t seconds after the start of the process are x and y respectively. It is given that :
= - 0.6 xy and x = 5 e-3 t^ , when t = 0 and y= 70 i) Form a differential equation in y and t. Solve this differential equation and obtain an expression for y in terms of t.
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A- LEVEL – MATHEMATICS P 3 EXCERCISE 1/ ANSWERS
Differential Equation
Q1. i) ln ( ) = - sin 4θ ii) x = 0.
Q7. i) = xy ii) y = 2
iii) Q2. i ) y = - ln ((1 – x )ex^ ) ii) ln (1-x) is def when 1-x > 0 or x<
Q8. Y =
Q3. y = 2x. (^) Q9. i) x = ii) K = ( 1- ) iii) x = [ t ∞ e-t^0 = 47.84 < 48 as t ∞ Q4. i) x = √( ) ii) x = or (2.60)
Q10. i) √M = 25 K sin (0.02 t)+ 10 ii) k =0. iii) M= [4.75 sin (0.02t)+10]^2 and least value of M = 27. Q5. i) y = ln [2 (tan x – x ) +1]
ii) y = 0.
Q11. i) = K (N-150) ii) N = 500 e0.08t^ + 150 iii) when t = 15, N = 1810 hence target of N= 2000 not met. Q6. i) ln( ) = t.ln3- ln ii) t = 4
Q12. y = 4 ( 2 + e3x^ )^2
Q13. i) = 0.02 N (1- 0.01 N)
ii) t = 50 ln ( ) iii) t = 49
Q19. i) K = 9 and x = [ 9 – 8 ii) when t is large.
Q14. x = ( 1 + tanθ)^2 - (^) Q20. ii) t = 60 [ 1- ]
iii) t = 49. Q15. i) R = x e( 3.80^ –^ 0.57x) or = 44.7xe-0.57x ii) R = 28.
Q21. ii) stationary points at and y = [ 9 e(2x + sin 2x)^ – At A when x= , y = 5. At B when x= , y = 48. Q16. i) y = [ x cos + sin + K ]^2
ii) K= y = [ x cos + sin + 10]^2 and y = 203 when x = 25
Q22. y = [ 2 e3x^ (3x-1) + And y = 2.44 when x= 0.
Q17 i) V = ( 80 – 80 e-kt) ii) K = 0. iii) V = 536.
Q23. y = ln [ ] Q24. i) y = 70 ii) y = Q25. y =
Q18. i) A=1 , B = - 2 , C =
ii) ln y = 1 - + 2 ln ( ) and y = when x = 2
Q26. y = ( x^2 + 4)^3
