Report

2nd Order Differential Equations Type 1 click for link Type 2 click for link Exceptions to the Particular Integral Rule 1st order equations which can be solved using the 2nd order method Equations which contain a equations. d2 y dx 2 term are called 2nd order differential Type 1 d2 y dy a 2 b cy 0 dx dx Step1 Define and solve the auxiliary equation al2 + bl + c = 0 Step 2 Write down the solution using the table below Roots of auxiliary equation Solution Real and different and y = Aex + Bex Real and equal (Repeated) y = (Ax + B)ex Complex roots i y = ex(Asinx + Bcosx) These solutions are called the complementary function The complementary function is the solution to the equation d2 y dy a 2 b cy 0 dx dx Ex1 Solve d2 y dy 4 3y 0 dx 2 dx Auxiliary equation 1l2 – 4l + 3 = 0 Solving (l 3)(l 1) 0 Ex2 Solve i.e and y = Ae3x + Be1x Complementary function solution Roots of auxiliary equation Real and different Real and equal (Repeated) l = 3 or 1 and Solution y = Aex + Bex y = (Ax + B)ex d2 y dy 4 4y 0 dx 2 dx Auxiliary equation 1l2 – 4l + 4 = 0 Solving (l 2)(l 2) = 0 Complementary function solution l=2 y = (Ax + B)e2x i.e repeated root Finding the two constants A 1st order differential equation contains one constant and a 2nd order contains two. Hence we need two boundary conditions. Ex1 d2 y dy 4 4y 0 2 dx dx Given that: when x = 0 y=8 Auxiliary equation l2 – 4l + 4 = 0 Solving (l 2)(l 2) = 0 Complementary function solution Now apply the boundary conditions and when dy 1 dx l=2 x=0 i.e repeated root y = (Ax + B)e2x y = (Ax + B)e2x a) Substitute x = 0 and y = 8 8 = Be0 B=8 b) Substitute dy = 1 when x = 0 dx dy = (Ax + B)2e2x + e2x(A) dx Using the product rule: 1stgrad 2nd + 2ndgrad 1st 1 = B(2e0) + Ae0 Substituting x = 0 and = 1 1 = 2B + A Substitute B = 8 so A = –15 The complete solution is therefore y = (Ax + B)e2x with A = –15 and B = 8 y = (–15x + 8)e2x Type 2 d2 y dy If the equation is of the form a 2 b cy f(x) dx dx then the solution is made up of the sum of the complementary function and a particular integral. The particular integral is determined using the table below. The particular integral is an expression which satisfies the differential equation when it is substituted in. f(x) linear polynomial ax + b exponential function aekx trig fn acospx or bsinpx or both Particular integral Cx + D Cekx Ccospx + Dsinpx F(x) Ex1 d2 y dy 4 2 4 y 3x 4 dx dx Solve 4l2 + 4l + 1 = 0 Auxiliary equation Solving (2l 1)(2l 1) = 0 l = – i.e repeated root Complementary function solution f(x) linear polynomial ax + b exponential function aekx trig fn acospx or bsinpx or both y = (Ax + B)e–x Particular integral Cx + D Cekx Ccospx + Dsinpx F(x) Particular integral is of the form y = Cx + D. This must satisfy the differential equation y = Cx + D dy C dx d2 y 0 2 dx d2 y dy 4 2 4 y 3x 4 dx dx d2 y dy 4 2 4 y 3x 4 dx dx y = Cx + D Substituting into the differential equation gives: 4(0) + 4C + Cx + D = 3x + 4 Comparing the coefficients x`s no.s C=3 4C + D = 4 D = –8 Thus the particular solution is y = 3x – 8 The complete general solution is given by: y = Complementary function + Particular integral y = (Ax + B)e–x + 3x – 8 dy C dx d2 y 0 2 dx Ex2 Solve d2 y dy 2 2 3y e2x dx dx Auxiliary equation Solving 2l2 – 1l – 3 = 0 (2l 3)(l 1) 0 l= Complementary function solution f(x) linear polynomial exponential function trig fn ax + b aekx acospx or bsinpx or both y = Ae or –1 x + Be–1x Particular integral F(x) Cx + D Cekx where k has the same value as the f(x) term Ccospx + Dsinpx Particular integral is of the form y = Ce2x. This must satisfy the differential equation d2 y dy 2 2 3y e2x dx dx d2 y dy 2 2 3y e2x dx dx y= Ce2x dy 2Ce2x dx d2 y 4Ce2x 2 dx Substituting into the differential equation gives: 24Ce2x – 2Ce2x – 3Ce2x = e2x Comparing the coefficients 8C – 2C – 3C = 1 e2x`s C = 1/3 Thus the particular solution is y = 1/3e2x The complete general solution is given by: y = Complementary function + Particular integral y = Ae x + Be–1x + 1/3e2x Solve d2 y 1 2 4y 3 sin x 4 cos x dx dy Here the term is missing so the auxiliary equation becomes dx 1l2 + 4 = 0 l = (0 + 2i) as –4 = 2i Complex roots Complementary function solution y = e0x(Asin2x + Bcos2x) = Asin2x + Bcos2x f(x) linear polynomial ax + b exponential function aekx trig fn acospx or bsinpx or both Particular integral Cx + D Cekx Ccospx + Dsinpx Particular integral is of the form y = Ccosx + Dsinx. y = Ccosx + Dsinx dy C sin x Dcos x dx d2 y Ccos x Dsin x 2 dx F(x) d2 y 1 2 4y 3 sin x 4 cos x dx dy C sin x Dcos x dx y = Ccosx + Dsinx d2 y Ccos x Dsin x 2 dx Substituting into the differential equation gives: –Ccosx – Dsinx + 4(Ccosx + Dsinx) = 3sinx + 4cosx 3Ccosx + 3Dsinx = 3sinx + 4cosx Equating coefficients of sinx and cosx 3D = 3 D=1 3C = 4 C= The complete general solution is given by: y = Complementary function + Particular integral y = Asin2x + Bcos2x + 1sinx + cosx Exceptions to the Particular Integral Rule If the particular integral is the same as one of the terms in the complementary function then use y = xf(x) as the particular integral. d2 y dy 2 2 3y e x dx dx 2l2 – 1l – 3 = 0 (2l - 3)(l + 1) = 0 l= or –1 Complementary function solution Auxiliary equation Solving y = Ae x + Be–1x As f(x) = e–x is contained in the complementary function solution use y = xCe–x as the P.I. d2 y dy 2 2 3y e x dx dx Particular integral is y = Cxe–x C.F. y = xCe–x y = Ae x + Be–1x P.I y = xCe–x as the f(x) term contains e–x as does the C.F dy Use the product rule to find dx and d2 y dx 2 dy Cx e x e x C = Ce−x (−x + 1) dx d2 y Ce x ( 1) ( x 1)( Ce x ) 2 dx Substituting into the differential equation gives: 2 Ce x ( 1) ( x 1)( Ce x ) Ce-x (-x +1) 3Cxe x e x 2 Ce x ( 1) ( x 1)( Ce x ) Ce-x (-x +1) 3Cxe x e x Comparing the coefficients –2C – 2C –1C e–x`s Thus the particular solution is =1 C=– y = – xe–x The complete general solution is given by: y = Complementary function + Particular integral y = Ae x + Be–x – xe–x Summary Complementary function contains Aex f(x) ie RHS of = Cex Particular Integral Dxex Acospx Ccospx Dxcospx Acoshpx Ccoshpx Dxcoshpx Rule Multiply the standard Particular Integral by x and then use dy d2 y and the product rule to find dx dx 2 1st order equations which can be solved using the 2nd order method dy b cy f(x) dx d2 y 0 can be treated as a 2nd order differential equation with dx 2 The following method only works if c is a number (not a P(x)) The solution will still consist of two parts: y = complementary function + particular integral dy 3y 2e x dx Auxiliary equation l–3=0 as l2 0 Complementary function solution y = Ae3x Particular integral is of the form y = Cex l=3 dy 3y 2e x dx P.I. y = Cex dy Ce x dx Substituting into the differential equation gives: Cex – 3Cex = 2ex –2C = 2 C = –1 The complete general solution is given by: y = Complementary function + Particular integral y = Ae3x – 1ex dy 3y 2e x dx Solution y = Ae3x – 1ex Compare this with the method using an integrating factor dy 3y 2e x dx I.F. e 3dx e 3x y I.F R.H.S I.Fdx y e-3x 2ex e3x dx 2e2x dx e2x C Multiply through by e3x y = Ce3x – 1ex As was obtained above