SECOND-ORDER LINEAR ODEs f +p(x)f +q(x)f = h(x) Generalities Structure of general solution Equations with constant coefficients
Second order linear equations General form : d f df + p( x) + q( x) f = h( x). Integrating factor? Suppose! I( x ) such that d If = Ih di d I = Ip and = Iq.. These equations are incompatible in most cases.
Structure of the general solution (GS) f +p(x)f +q(x)f = h(x) The general solution f is the sum of a particular solution f 0 (the particular integral, PI) and the general solution f 1 of the associated homogeneous equation (the complementary function, CF): f = f 0 +f 1, i.e., GS = PI + CF. The complementary function CF is given by linear combination of two linearly independent ( see next) solutions u 1 and u : CF = c 1 u 1 (x)+c u (x) c 1 and c arbitrary constants
Two functions u 1 (x) and u (x) are linearly independent if the relation αu 1 (x)+βu (x) = 0 implies α = β = 0. Let αu 1 (x)+βu (x) = 0. Differentiating αu 1(x)+βu (x) = 0. If W(u 1,u ) = u 1 u u 1 u = u 1u u u 1 0 then α = β = 0, and u 1 and u are linearly independent. If W(u 1,u ) = u 1 u u u 1 = 0 then u = constant u 1 u 1 and u not linearly independent W(u 1,u ) = wronskian determinant of functions u 1 and u n functions u 1 (x),...,u n (x) are linearly independent if α 1 u 1 (x)+...+α n u n (x) = 0 = α 1 =... = α n = 0.
Example: The functions u 1 (x) = sinx and u (x) = cosx are linearly independent. Let αsinx+βcosx = 0. Differentiating αcosx βsinx = 0. So α = β sinx cosx β ( sin ) x cosx +cosx = 0 β 1 cosx = 0 β = 0. Thus α = β = 0. Alternatively: W(u 1,u ) = u 1 u u u 1 = sin x cos x = 1 linear independence
Homework Determine whether the following sets of functions are linearly independent. x, x, 1 [Answ.: linearly independent] 1 x, 1+x, 1 3x [Answ.: linearly dependent] e x, e x [Answ.: linearly independent] e x, xe x, x e x, x 3 e x [Answ.: linearly independent] sinx, cosx, sin(x+α) [Answ.: linearly dependent]
Homogeneous equation: f +p(x)f +q(x)f = 0 Any solution u can be written as a linear combination of linearly independent solutions u 1 and u. Since u, u 1 and u all solve the homogeneous eq., with nonzero coefficients of the second-derivative, first-derivative and no-derivative terms, we must have det = 0 u u 1 u u u 1 u u u 1 u = 0 W(u,u 1,u ) = 0 for solutions u, u 1, u αu+βu 1 +γu = 0 for α, β and γ not all zero. Solving for u expresses the solution u as a linear combination of u 1 and u. CF = c 1 u 1 (x)+c u (x) solutions span whole set of linear combinations of two independent u 1, u
Example: The general solution of the nd-order linear ODE y +y = 0 is Asinx+Bcosx. (simple harmonic oscillator) To show this, it is sufficient to show that i) sinx and cosx solve the equation (e.g. by direct computation) and ii) sin x and cos x are linearly independent (see previous Example). Then general theorem CF = c 1 u 1 (x)+c u (x) yields the result.
Useful reference for this part of the course, with worked problems and examples, is Schaum s Outline Series Differential Equations R. Bronson and G. Costa McGraw-Hill (Third Edition, 006) See chapters 8 to 14.
nd-order linear ODEs with constant coefficients: general methods of solution available arise in many physical applications Second order linear equation with constant coefficients d f df Lf = a a 1 a0f h( x) + + =.
Second order linear equation with constant coefficients Solution d f df Lf = a a 1 a0f h( x) + + =. Complementary function The number of independent complementary functions is the number of integration constants equal to the order of the differential equation 1) Construct f the general solution to the homogeneous equation Lf = 0 0 0 ) Find a solution, f, to the inhomogeneous equation Lf 1 1 = h Particular integral General solution : f0 + f1 For a nth order differential equation need n independent solutions to Lf=0 to specify the complementary function
Second order linear equation with constant coefficients d f df Lf = a a 1 a0f h( x) + + =. Complementary function d f df Lf = a a 1 a0f 0 + + =. Try y = e mx am + am+ a =. 1 0 0 m ±! a1 ± a1! 4aa 0 ", a Auxiliary equation a! 4 aa " +,0,! 1 0 Complementary function m x m x y = Ae + + Ae!. +! Two constants of integration
Ex 1 d y dy + 4 + 3y = 0. Auxiliary eq. ( m+ 3)( m+ 1) = 0 " CF is y = Ae + Be! 3x! x. If y(0)=, y'(0)=0! A+ B=, " 3A" B= 0 Initial conditions A=-1, B=3 " y =! +! 3x! x e 3e
Ex Ly = d y dy 5y 0 + + =.. Auxiliary eq. m + m+ 5=. 0 Complex? i.e. m= (! ± 4! 0) =! 1 ± (! 1+ i) x (! 1! i) i " CF is y = Ae + Be 1 x But Lis a real operator! 0=" e( Ly) = L[ " e( y)]! x ie..! e( y) is a solution (as is " m( y)) $ # e( y) = e [ A" cos( x) + B" sin( x)]. Find the solution for which y (0) = 5 and (dy/ d x) 0 = 0 Initial conditions 5= A" 0=! A" + B" # B" = # y = e [5cos( x) + sin( x)]. 5! x 5
Factorisation of operators We wish to solve d f df Lf = a a 1 a0f 0 + + =. We did this by trying y = e mx am + am+ a = a ( m! m )( m! m ) = 0. 1 0 +! m ±! a ± a! 4aa = a 1 1 0 This is equivalent to factorising the equation d d d f df!!! + =!! + + +! + ( m )( m ) f ( m m ) mm f d f a df a Lf 1 0 = + + " a a a Now we can see why the CF is made up of exponentials. because : d m d! x m+ x (! m! )e = 0 ; (! m+ )e = 0.
Factorisation of operators and repeated roots d f df Lf = a a 1 a0f 0 + + =. d d (! m! )(! m+ ) f = 0 m ±! a ± a! 4aa = a 1 1 0 What happens if a! 4aa = 0 1 0 and m = m = m? +! Lf = d d ( )( )! m! m f. mx mx d d mx e gives one solution : L(e ) = ( m)( m)e 0!! = mx xe gives the second independent solutio n: mx d d mx d mx L( xe ) = (! m)(! m) xe = (! m)e = 0, mx y = Ae + Bxe mx
Ex 4 d y dy + + y = 0. Auxiliary equation ( m+ 1) = 0 x y = Ae + Bxe!! x. Ex 5 4 d 3 d d d 4 d 3 d d d y y y y! +! + y = 0. x x x x Auxiliary equation ( m! 1) ( m! i)( m+ i) = 0 x y = e ( A+ Bx) + Ccosx+ Dsinx.
Summary nd order linear ODEs: f +p(x)f +q(x)f = h(x) General solution = PI + CF CF = c 1 u 1 +c u u 1 and u linearly independent solutions of the homogeneous equation Equations with constant coefficients: Solve auxiliary equation to find complementary function CF distinct real roots repeated roots complex roots Next: methods to find the particular integral PI