Understanding Second Order Homogeneous Differential Equations

Differential equations In this lecture we will: Look at second order homogeneous differential equations. Introduce the auxiliary equation and determine its roots. Find out how to solve the homogeneous second order differential equation in the case that the roots of the auxiliary equation are: Real and different. The same. Complex conjugate. Some comprehension questions for this lecture. Write down the general form of a homogeneous second order differential equation with constant coefficients. Solve the initial value problem: 2 d y dy 4 4y dx dx + = 0, 2 dy(1) dx = = with y(1) 1 and 1 1

Homogeneous second order differential equations Consider second order homogeneous differential equations of the form: d y dy a b cy dx dx Substituting into the original equation we have: 2 mx mx am e bme or e (am bm + + + = mx ce c) + 0, 2 + + = 0. 2 = mx 2 0 The coefficients a, b and c are all constants. Try to find a solution of the form mx y e . = Now emx cannot be zero, so: 2 am bm + + = c 0. This is called the auxiliary equation. The above implies that y = emx is a solution of the differential equation iff (if and only if) m takes one of the values: 2 b b m 2a b b m 2a Differentiating this gives: dy me dx 2 d y dx = = mx 2 mx and m e . 2 + 4ac = , 1 2 4ac = . 2 2

Homogeneous second order differential equations 2 When the discriminant m1 and m2 are real and distinct. When the roots are real and equal. When the roots are complex conjugate numbers. Prove this: d a dx b 4ac 0, 2 ( ) + + C y C y = 2 1 1 2 2 b 4ac 0, 2 d ( ) ( ) + + + b C y C y c C y C y 2 b 4ac 0, 1 1 2 2 1 1 2 2 dx d y dx 2 2 d y dx = + + 1 2 aC aC 1 2 2 2 The principle of superposition: Supposing we have two solutions of our homogeneous second order differential equation, y1(x) and y2(x). The sum C1y1(x) + C2y2(x) is also a solution of the equation. dy dx dy dx + + + 1 2 bC bC cC y cC y 1 2 1 1 2 2 2 d y dx dy dx = + + + 1 1 C a b cy 1 1 2 2 d y dx dy dx + + 2 2 C a b cy 2 2 2 = 0. 3

Homogeneous second order differential equations Consider various possibilities for the solutions of the auxiliary equation. If we have distinct roots, and are linearly independent solutions of our differential equation. Example: Find a general solution of: 2 d y dy 5 dx dx = m x 1y e 1 + = 6y 0. = m x 2 y e 2 2 The auxiliary equation is: 2 m 5m 6 (m 6)(m 1) m 1and m = + = 0 + = = 0 The functions y1(x) and y2(x) are linearly independent if one is not just a multiple of the other, that is: y (x) ky (x). 6. 1 2 Hence a general solution to the equation is: x y(x) Ae Be = + 2 1 6x . Hence, by the superposition principle, a general solution is: m x y(x) Ae = + m x Be . 1 2 4

Homogeneous second order differential equations Another example, solve the initial value problem d y dy 2 y dx dx y(0) dy(0) dx A general solution is y(x) Ae = ( ) ( ) + 1 2 x 1 2 x + Be . 2 The initial conditions can be used to determine A and B: y(0) Ae Be 0 A B or A dy A 1 2 e dx dy(0) 1 2 A dx 1 1 2 A = + + = 0 2 = 0 = = + 0 0 + = B 1 or y (0) = = 1. ( ) ( ) ( ) ( ) + 1 2 x 1 2 x = + + B 1 2 e The auxiliary equation is: m 2m 1 + = + = ( ) ( ) 2 0 = + + 1 2 B 2 8 ( ) ( ) = + m 1 2 1 2 A 1 2 2 8 1 = = and m 1 2. = = 1 2 2A or A . 2 2 2 2 5

Homogeneous second order differential equations Rewriting: If the roots of auxiliary equation are the same (m), we can use y = emx and y = xemx as two linearly independent solutions of the differential equation. Example: Find a general solution of: 2 d y dy 4 4y dx dx 2 2 = = A and B . 4 4 Putting this together: ( ) ( ) 2 2 + 1 2 x 1 2 x = + y(x) e e . 4 4 + + = 0. 50 2 y(x) 40 Auxiliary equation: 2 m 4m (m m m = 30 + + = 4 0 yda xa ( ) 20 + = 2 2) = 0 10 2. 1 2 0 10 x 2 1 0 1 2 xa 6

Homogeneous second order differential equations General solution therefore: 2x y(x) Ae = Writing P = A + B and Q = i(A B), we then have the general solution: ( y(x) e Pcos( x) Qsin( x) . = 2x + Bxe ) + x Example: Find the general solution of: 2 d y dy 2 dx dx What if the auxiliary equation has complex conjugate roots m i and m = + = i ? + + = 4y 0. 1 2 2 Then: y ( ) ( ) + i x i x Auxiliary equation: m 2m + + = = = + Ae Be + ( ) + = 2 4 0 i x i x x e Ae Be 2 4 16 2 2 4 16 2 3 ( ) A cos( x) isin( x) + + = m , m = x 1 2 e ( ) x) isin( + B cos( x) = + = m 1 3, m 1 1 2 + + (A B)cos( x) i(A = 1 i 3 and m = + 1 i 3. = x e . or m 1 2 B)sin( x) 7

Homogeneous second order differential equations = cos 3x = 1 and + 3: So, with y = Using the initial conditions we have: y(0) P 1, dy(0) 2Q 0, so Q dx = = x x Pe Qe sin 3x. Another example: Solve the initial value problem: 2 d y 4y 0, y(0) dx = = = 0. The required solution is therefore: y(x) cos2x. = dy(0) dx + = = = 1, 0. 2 Auxiliary equation: 2 m 4 m = = + = 0 = 2i, m 0, 2i. 1 2 = Hence This gives: y dy dx 2. = + Pcos2x Qsin2x, 2Psin2x = + 2Qcos2x. 8