SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS

We shall consider solutions to equations of the form: d2 y d x2 dy d x a + b + cy = f(x) ay + by + cy = f(x) or: We shall see that the solution consists of two parts, the complementary function (C.F.) which is the solution to The equation when f(x) = 0, and the particular integral, (P.I.) which relates to f(x) when f(x) ≠ 0.

The complementary function If we guess that a solution of ay + by + cy = 0 is y = eαx with α to be determined: So: y = eαx y = αeαx y = α2eαx Substitute these into the original equation: Then: aα2eαx + bαeαx + ceαx = 0 eαx ( aα2 + bα + c ) = 0 Hence y = epx and y = eqx , where p and q are the roots of aα2 + bα + c = 0 , are solutions. i.e. The C.F. is y = Aepx + Beqx.

We must now consider the two special cases which arise when aα2 + bα + c = 0 ( called the auxiliary equation ) has either Two equal roots, or ii) Complex roots. i) If p = q By observing the sum and product of the roots: = – 2p, and b a c = p2 The differential equation can then be written: y – 2py + p2y = 0 … (1) Let y = epxg(x), where g(x) is some function of x. y = e pxg (x) + g(x)pe px y = e pxg (x) + 2pg (x)e px + p2g(x)e px In (1): e px[ g (x) + 2pg (x) + p2g(x) – 2pg (x) – 2p2g(x) + p2g(x) ] = 0 e pxg (x) = 0 g (x) = 0 g(x) = Ax + B Hence the C.F. is: y = epx (Ax + B )

ii) If p and q are complex, we can write: p = a + ib, q = a – ib. The C.F. is y = Ae(a + ib)x + Be(a – ib)x = Aeaxeibx + Beaxe–ibx = eax( Aeibx + Be–ibx ) = eax [A(cosbx + isinbx) + B(cos(–bx) + isin(–bx))] = eax [A(cosbx + isinbx) + B(cosbx – isinbx)] = eax [Ccosbx + Dsinbx] ( where C and D are constants, which include complex values)

The solutions to the differential equation: d2 y d x2 dy d x a + b + cy = 0 depend on the roots of the auxiliary equation: au2 + bu + c = 0 If b2 – 4ac > 0 and has roots p and q y = Aepx + Beqx If b2 – 4ac = 0 with repeated root p y = epx (A + Bx) If b2 – 4ac < 0 with complex roots p  iq y = epx ( Acosqx + Bsinqx )

Obtain the general solutions of the following equations: Example 1: Obtain the general solutions of the following equations: i) 4y – y – 5y = 0 ii) y – 4y + 13y = 0 i) The auxiliary equation is: 4u2 – u – 5 = 0 – 1 u = 5 4 , Factorising this: ( 4u – 5 ) ( u + 1 ) = 0 y = Ae + Be – x 5x 4 The general solution is: ii) The auxiliary equation is: u2 – 4u + 13 = 0 = 2 ± 3i The general solution is: y = e2x ( Acos3x + Bsin3x)

d2 y d x2 dy d x – 6 + 9y = 0 Example 2: Solve the differential equation: dy d x = 7. given that when x = 0, y = 1 and The auxiliary equation is: u2 – 6u + 9 = 0 ( u – 3 ) ( u – 3 ) = 0 u = 3 ( repeated root ) The general solution is: y = e3x (Ax + B) …(1) dy d x = Differentiating (1): e3xA + (Ax + B)3e3x …(2) Now, if x = 0, y = 1 in (1): 1 = 1 ( 0 + B ) so, B = 1 dy d x = 7 in (2): Also,if x = 0, 7 = 1 ( A ) + ( 0 + 1 ) 3 so, A = 4 The solution is: y = e3x (4x + 1 )

The particular integral when f(x) ≠ 0. Now, returning to the equation: d2 y d x2 dy d x a + b + cy = f(x) To establish the particular integral (P.I.), certain expressions for f(x) suggest standard trial solutions: f(x) Trial solution p y = λ px + q y = λx + μ psinx + qcosx y = λsinx + μcosx e kx y = λe kx The solution to the differential equation is then y = C.F + P.I.

Example 3: Find the general solution of the differential equation: d2 y d x2 dy d x – 2 – 3y = e 2x The auxiliary equation is: u2 – 2u – 3 = 0 ( u – 3 )( u + 1) = 0 u = 3, – 1 The C.F. is: y = Ae3x + Be – x dy d x = d2 y d x2 = For the P.I. try y = λe2x 2λe2x 4λe2x Substitute these into the original equation: λ = 1 3 – 4λe2x – 2(2λe2x) – 3(λe2x) = e2x y = Ae3x + Be – x 1 3 e2x – The general solution is:

Obtain the general solutions of the following equation: Trials that fail Example 4: Obtain the general solutions of the following equation: y – 3y + 2y = e2x. The auxiliary equation is: u2 – 3u + 2 = 0 ( u – 1 )( u – 2) = 0 u = 1, 2 The C.F. is: y = Aex + Be 2 x For the P.I. we would normally try y = λe2x, but this is already part of the solution. Try y = λ x e2x y = λx2e2x + e2x λ = λe2x ( 2x + 1 ) y = λe2x 2 + ( 2x + 1 )2λe2x = 4λe2x ( x + 1 ) Sub. into the original equation: 4λe2x ( x + 1 ) – 3 λ e2x ( 2x + 1 ) + 2 λ x e2x = e2x λe2x = e2x λ = 1 The general solution is: y = Aex + Be 2 x + xe2x

The solution to the differential equation: d2 y d x2 dy d x a + b Summary of key points: The solution to the differential equation: d2 y d x2 dy d x a + b + cy = f(x) is y = C.F + P.I. The C.F. is found from the auxiliary equation: au2 + bu + c = 0 If b2 – 4ac > 0 and has roots p and q, y = Aepx + Beqx If b2 – 4ac = 0 with repeated root p, y = epx (A + Bx) If b2 – 4ac < 0 with complex roots p  iq, y = epx(Acosqx + Bsinqx) To establish the P.I. the given trial solutions are possible. f(x) Trial solution p y = λ px + q y = λx + μ psinx + qcosx y = λsinx + μcosx e kx λe kx If the trial solution is part of the C.F. then try multiplying by x. ( If this also fails, try multiplying by x2 ). This PowerPoint produced by R.Collins ; Updated Feb.2010