RI Differential Equations Solns

RAFFLES INSTITUTION H2 Mathematics 9758 2023 Year 6 Term 3 Revision 12 (Summary and Tutorial) Topic: Differential Equations Summary for Differential Equations Differential equations of the form d f( )d y xx = - solved by “direct” integration, i.e. integrate f( )x with respect to x. d f( ) f( ) dd y x y xxx = ⇒= ∫ Example [9740/2008/01/Q4(i)(ii))] (i) Find the general solution of the differential equation 2 d3 d1 yx xx= + . (ii) Find the particular solution of the differential equation for which 2y= when 0x= . Solution: (i) 2 d3 d1 yx xx= + 22 2 2 33 d [integrate the f( ) with respect to ]11 32 d21 3ln( 1)2 xxyx x xxx x xx xc = =++ = + = ++ ∫ ∫ (ii) When 0, 2xy= = 232 ln(0 1)2 c= ++ 2c= Thus the particular solution is 23ln( 1) 22yx= ++ .

Differential equations of the form d f( )d y yx = - solved by separating the variables, i.e. d1 f ( ) d 1 dd f( ) y y yxxy= ⇒= ∫∫ Example [9740/2012/02/Q1(b)] Given that u and t are related by 2d 16 9d u ut = − and that 1u= when 0t = , find t in terms of u, simplifying your answer. Solution: 2 2 22 d 16 9d 11 d 1 d [ separate the variables to obtain d 1 d ]16 9 f ( ) 13 d 1 d3 4 (3 ) 1 1 43 ln3 2(4) 4 3 1 43ln24 4 3 u ut ut utuu utu u tcu u tcu = − = =− =− +×= +− + = +− ∫ ∫ ∫∫ ∫∫ When 0, 1tu= = 1 ln724c= Thus 1 43 1 1 43ln ln 7 ln24 4 3 24 24 7(4 3 ) uut uu ++= −= −−

Differential equations of the form 2 2 d f( )d y xx = - solved by integrating the right hand side twice with respect to x. Note that 2 2 dd f( ) f( ) ddd yy x xxxx = ⇒= ∫ Now suppose that f () d F ()xx xc= +∫ , then we have d F( )d y xcx = + which gives F( ) dy x cx= +∫ Example [9740/2012/02/Q1(a)] Find the general solution of the differential equation 2 2 2 d 16 9d y xx = − Giving your answer in the form f( )yx= . Solution: 2 2 2 2 3 3 24 d 16 9d d 16 9 d [integrate with respect to ]d 16 3 16 3 d [integrate with respect to ] 38 [note that there are two constants and ]4 y xx y xx xx xxc y x x cx x x x cx d c d = − = − =−+ = −+ = − ++ ∫ ∫

Solving Differential Equations using a given substitution - the substitution will be given in the question - this substitution allows us to reduce the differential equation to familiar forms which we discussed earlier, namely d f( )d y xx = or d f( )d y yx = . Example [9233/N87/02/Q14(a)] The variables x and y are connected by the differential equation d1 d1 y xy x xy ++= −− . Show that the substitution uxy= + reduces the equation to d2 d1 u xu= − , and solve this differential equation. Solution: dd 1 [ differentiating both sides with respect to ]dd uyuxy xxx=+⇒ =+ dd 1dd yu xx∴=− d1 d1 d1 1d1 d1 1d1 11 1 2 (shown)1 y xy x xy uu xu uu xu uu u u ++= −− +−= − += +− ++−= − = − 2 (1 ) d 2 d [separate the variables] 1 22 uu x u u xc −= −= + ∫∫

Differential Equations in Real-World Context - this includes formulating the differential equation in the given context, solving the equation and interpreting the solution of the differential equation. In general, one can use Rate of change Rate increase Rate decreas e= − to formulate the differential equation for a given real-world context problem. Example [9740/2010/01/Q7] A bottle containing liquid is taken from a refrigerator and placed in a room where the temperature is a constant 20 C° . As the liquid warms up, the rate of increase of its temperature θ C° after time t minutes is proportional to the temperature difference (20 )θ− C° . Initially the temperature of the liquid is 10 C° and the rate of increase of the temperature is 1 C° per minute. By setting up and solving a differential equation, show that 1 1020 10 t eθ − = − . Find the time it takes the liquid to reach a temperature of 15 C° , and state what happens to θ for large values of t. Sketch a graph of θ against t. Solution: d (20 )d kt θ θ= − [rate of change d dt θ= , rate increase (20 )k θ= − where k is the constant of proportionality] When d10, 1dt θθ = = 11 (20 10) 10kk= − ⇒= 11 d d20 10 tθθ =−∫∫ [solving DE of the form d f( )d y yx = ] 1 10 1 10 1 10 1ln 20 10 20 20 w here 20 tc t c t tc e Ae A e Ae θ θ θ θ −− − − − − −= + −= ± −= = ± = − When 0, 10t θ= = 10 20 10 AA= −⇒= formulating interpreting

1 1020 10 t eθ − ∴= − When 15θ = , 1 10 1 10 15 20 10 1 2 110ln 6.93 mins2 t t e e t − − = − = = −= As t→∞ , 1 10 0 t e − → and so 20θ → . Example [9233/N74/02/Q20] A race called the Matrices live on an isolated island called Vector. The number of births per unit time is proportional to the population at any time and the number of deaths per unit time is proportional to the square of the population. If the population at time t is p, show that 2d d p ap bpt = − , where a and b are positive constants. Solve the equation for p in terms of t, given that 2 3 ap b= when 0t = . Show that there is a limit to the size of the population. Solution: Rate increase p∝ Rate increase ap∴= where a is a positive constant Rate decrease 2 p∝ 2 Rate decrease bp∴ = where b is a positive constant Thus 2d d p ap bpt = − [formulating] 2 1 d 1 dptap bp =−∫∫ [solving DE of the form d f( )d y yx = ] … Refer to Q12 of your C9 Differential Equations tutorial for the remaining parts of the solution. t 0

Revision Tutorial Questions Source of Question: ACJC JC2 CT1 9758/2018/Q2 1 In a factory process, a chemical C is used up at a rate proportional to the square of the amount of chemical present, x grams, at time t seconds. At the start of the process, 500 grams of C is present. By setting up and solving a differential equation, express the solution of the differential equation in the form f( )xt= , and sketch the part of the curve with this equation which is relevant in this context. [6] [ 500 1 500x kt= − ] 1 [6] 2d d x kxt = where k is a negative constant 2 1 dd 1 x kt x kt cx ⇒= ⇒− = + ∫∫ When t = 0, x = 500, 1 500 c−= 1 500 1 500 1 500 x ktkt −−⇒= = −− or 500 1 500x kt= − Any graph with negative value for k. 500 O t x

Source of Question: DHS JC2 Prelim 9758/2018/01/Q6 2 (a) By using the substitution 2 ,y zx= find the general solution of the differential equation 22d 2 , where 0.d yx xy y xx = −≠ [4] (i) Sketch the solution curve that passes through ( )2, 4 ,− indicating any stationary points and asymptotes clearly. [4] (ii) State the particular solution for which y has no turning point. [1] (b) A differential equation is of the form d ,d y y px qx+= + where p and q are constants. Its general solution is 4 1 e, xyx D −= −+ where D is an arbitrary constant. Find the values of p and q. [2] [(a) 2xy xC= − (a)(ii) yx= (b) 4, 3pq= = ] 2(a) [4] 22d 2d yx xy yx = − ...(1) Differentiating both sides of 2y zx= with respect to x: 2d dz2dd y xz xxx= + ...(2) Substitute (2) into (1): ( ) ( ) 22 2 22 4 24 d22 d d d zx xz x x zx zx x zx zxx  += − =− 2 2 2 2 d d 1 d 1d 1 z zx zxz xCz x xCy xy xC =− =− − = −+ − = −+ = − ∫∫

2(a)(i) [4] Given ( )2, 4 ,− 2243 2 CC−= ⇒ =− Hence, 2 9333 xyx xx= =++−− 2(a)(ii) [1] When 0,C = particular solution is 2 0 xyx x= =− which is a straight line and has no turning point. 2(b) [2] Given d41 e 4 e d xx yyx D D x −−= −+ ⇒ =− ( ) ( )d 4 e 41 e 43d xxy y D xD xx −−+ = − + −+ = + 4, 3pq∴= = Source of Question: NJC JC2 Midyear 9758/2018/Q2 3 (i) Show that the substitution 222= +rxy reduces the differential equation 22d 4d =−− −yy xx yx to 2d 4.d = −rrr x Hence solve the differential equation, given that y = 2√2 when 0.=x [7] The solution to part (i) is the equation of a curve C. (ii) Explain why (a) every point on C is more than 2 units away from the origin O. [1] (b) C is symmetrical about the x-axis. [1] [ ( ) 22 2 41 e xxy −+= + ] y x y = x