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