F03 - Further Differential Equations - Lecture Notes (Teacher s Version)

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Teacher’s Version) Page 1 of 42 National Junior College 2016 – 2017 H2 Further Mathematics Topic F3: Further Differential Equations (Lecture Notes) Key Questions to Answer: 1. How do we solve differential equations of the following forms analytically? (i)    d f gd y x yx  , (ii) d P( ) Q( ).d y x y xx  , (iii) 2 2 d d 0d d y ya b cyx x   , (iv)   2 2 d d fd d y ya b cy xx x   , where  f x is a polynomial or ekxp or    cos sinp kx q kx  , including equations that can be reduced to the above by means of a given substitution. 2. How do we sketch a family of solution curves of a differential equation? 3. How do we determine the equilibrium points and draw the phase lines of autonomous differential equations? 4. How do we model and solve problems related to the spread of diseases or population growth, with competition and harvesting? 5. What is the relationship between the solution of a nonhomogeneous equation and the associated homogeneous equation? 6. How do we model and solve problems related to the motion of particles that involves resistance, free or driven oscillation and damping? §1 Analytic Solutions of First Order Differential Equations 1.1 Separation of Variables A first-order differential equation of the form d f ( )g( )d y x yx  is said to be separable or to have separable variables. www.KiasuExamPaper.com 192

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Teacher’s Version) Page 2 of 42 Example 1.1.1 Solve d(1 ) d yx y x  , expressing y in terms of x. Solution: d(1 ) d yx y x  d d 1 y x y x      ln ln 1y x C    ln 1 y Cx   e1 Cy x  e1 Cy x  , e1 Cy A Ax   (1 )y A x   Example 1.1.2 Solve 2d 4d y yx   , expressing y in terms of x. Solution: 2d 4d y yx   2 d 1 d4 y xy     2 2 d 2 y x Cy    1 2ln2(2) 2 y x Cy    2ln 4 2 y x Cy     4 42 e e , e2 x C x Cy A Ay      42 e ( 2) xy A y    4 4e 2 e 2x xy Ay A    4 4(1 e ) 2(1 e )x xy A A      4 4 2 1 e 1 e x x A y A     www.KiasuExamPaper.com 193

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Teacher’s Version) Page 3 of 42 Example 1.1.3 The gradient of a curve at any point ( x, y) is given by the expression d , 0.d y y y xx x   Find the equation of the curve if it passes through the point   22, e . Solution: d 11d y y y yx x x         d 1 1 dy xy x     ln lny x x C    ln xy x C   ex Cxy   e e , ex C x Cxy A A     exAy x  Since the curve passes through   22, e , 2 2 ee 22 A A   2ex y x  Example 1.1.4 Find the general solution of the differential equation d 2 d y x x y . Sketch 3 members of the family of solution curves. Solution: d 2 d y x x y d 2 dy y x x   2 2 2 y x C  2 2 12 y x C C  Note: Family of solution curves is a set of ellipses. 2 ̶ -2 www.KiasuExamPaper.com 194

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Teacher’s Version) Page 4 of 42 Example 1.1.5 Show that the differential equation 2 2d2 2d yxy x yx   can be reduced by means of the substitution y vx to d2 1 d vvx x  . Hence, find the general solution, giving 2y explicitly in terms of x. Solution: d d d d y vy vx v x x x    2 2d2 ( ) 2( )d vx vx v x x vx x        d2 1 d vvx x  12 d dv v x x 2 2 2 2 2 2ln ln e y C xyv x C x v C C x x            2 2 2 2 2 2 2 2 1e e , e e ln( ) , y y yC Cx x xx yx A A Bx BA x A           2 2 ln( )y x Bx  1.2 Linear Equations and Integrating Factor A linear differential equation takes the general form ( ) ( 1) 1 0 ( ) ( ) ... ( ) g( )n n n na x y a x y a x y x      , where ( ) d d n n n yy x . Every first-order linear differential equation can be expressed in the standard form d P( ) Q( ).d y x y xx   For a first-order differential equation of the above form, the function P( ) d I( ) e x x x  is defined to be the integrating factor of the differential equation. The following proof illustrates how the integrating factor can be used to solve a linear first -order differential equation of the form d P( ) Q( ).d y x y xx   Proof: Multiplying both sides of the differential equation by P( ) d I( ) e , x x x  we obtain www.KiasuExamPaper.com 195

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Teacher’s Version) Page 5 of 42 P( ) d P( ) d P( ) d P( ) d P( ) d P( ) d P( ) d P( ) d P( ) d P( ) d P( ) d P( ) d P( ) d de P( ) e Q( )d de e P( ) e Q( )d d de e e Q( )d d d e e Q( )d e e Q( ) d e x x x x x x x x x x x x x x x x x x x x x x x x x x y x y xx y y x xx y y xx x y xx y x x y                               P( ) d P( ) d Q( ) d I( )Q( ) d , where I( ) e . I( )e x x x x x x x x x xx     This procedure in solving a differential equation is called the method of integrating factor. Example 1.2.1 Solve the differential equation 2d 3 5d yx y xx   for 0x , expressing y in terms of x. Solution: Step 1: Rewrite the differential equation in standard form. 2d d 33 5 5d d y yx y x y xx x x          Step 2: Compute the integrating factor.   3 d 3ln 3e e x xxI x x    Step 3: Multiply both sides of the differential equation (in standard form) by the integrating factor. 3 2 4d 3 5d yx x y xx   Step 4: Rewrite the equation into an exact differential.   3 4d 5d x y xx  Step 5: Solve the differential equation by direct integration. 3 4 3 5 2 3 5 dx y x x x y x C Cy x x       Will the solution be different if 0x ? www.KiasuExamPaper.com 196

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Teacher’s Version) Page 6 of 42 Example 1.2.2 Express x in terms of t given that 3d d xt x tt   and x = 1 when t = 1. Solution: 3d d xt x tt   2d 1 d x x tt t       Integrating factor 1 1d ln ln 1e e e t t tt t       2 3 2 1 d 1 d dd d 2 2 x x x x t tx t t t t C x Ctt t t t t t t                        At t = 1, x = 1, 1 11 2 2C C    3 2 2 t tx   . Example 1.2.3 Find the solution of the differential equation 3dsin cos sind yx y x xx   , given that y = 1 when x = π 2 . Solution: 3 2d d cossin cos sin sind d sin y y xx y x x y xx x x          Integrating factor cos d ln(sin )sin 1e e sin x x xx x      2 1 d cos sinsin d sin y x y xx x x             d sind sin sin dsin cossin y xx x y x xx y x Cx              At x = π 2 , y = 1,  π 2 1 πcos 1sin 2 C C       sin (1 cos )y x x   www.KiasuExamPaper.com 197

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Further Differential Equations (Teacher’s Version) Page 7 of 42 Example 1.2.4 By means of the substitution 2 1 2zy  , show that the differential equation 3d d y y yx x  may be reduced to d 2 1d z z x x  . Hence find the solution of 3d ,d y y yx x  given that y = 2 when x = 1. Solution: 2 3 1 2 d d2 2d d y zzy y x x   3d d d d y z yx x  3 3 3d d d d y y z yy y yx x x x      2 d 1 1d z x xy   d 2 1d z z x x   (shown) Integrating factor = 2 d 2ln 2 1e e x xx x      2 3 2 1 d 2 1 d z z x x x x   2 2 d 1 d z x x x      2 2 1 dz xx x   2 1z C