1. 2 nd Order Differential Equations Type 1 click for link Type 2click for link Exceptions to the Particular Integral Rule 1 st order equations which can be solved using the 2 nd order method

2. Step1 Define and solve the auxiliary equation a 2 + b + c = 0 Step 2 Write down the solution using the table below Roots of auxiliary equationSolution Real and different  and  y = Ae  x + Be  x Real and equal (Repeated)  y = (Ax + B)e  x Complex roots  i  y = e  x (Asin  x + Bcos  x) These solutions are called the complementary function The complementary function is the solution to the equation Equations which contain a term are called 2nd order differential equations. Type 1

3. Ex1 Solve Auxiliary equation 1 2 – 4 + 3 = 0 Solving (  0  = 3 or 1 i.e  and  Complementary function solution y = Ae  x + Be  x Ex2Solve Auxiliary equation 1  – 4 + 4 = 0 Solving (  = 0 = 2i.e  repeated root Complementary function solution y = (Ax + B)e  x Roots of auxiliary equationSolution Real and different  and  y = Ae  x + Be  x Real and equal (Repeated)  y = (Ax + B)e  x

4. 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 Given that: when x = 0y = 8 and when x = 0 Auxiliary equation  – 4 + 4 = 0 Solving (  = 0 = 2i.e  repeated root Complementary function solutiony = (Ax + B)e  x Now apply the boundary conditions

5. a) Substitute x = 0 and y = 8 8 = Be 0 B = 8 b) Substitute = 1 when x = 0 = (Ax + B)  2e 2x + e 2x (A) Using the product rule: 1st  grad 2nd + 2nd  grad 1st 1 = B  (2e 0 ) + Ae 0 Substituting x = 0 and = 1 1 = 2B + A Substitute B = 8so A = –15 The complete solution is therefore y = (Ax + B)e 2x with A = –15 and B = 8 y = (–15x + 8)e 2x y = (Ax + B)e  x

6. Type 2 If the equation is of the form 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)Particular integral F(x) linear polynomial ax + bCx + D exponential functionae kx trig fnacospx or bsinpx or both Ce kx Ccospx + Dsinpx

7. Ex1Solve Auxiliary equation 4 2 + 4 + 1 = 0 Solving (  = 0 = –  i.e  repeated root Complementary function solutiony = (Ax + B)e –  x Particular integral is of the form y = Cx + D. This must satisfy the differential equation y = Cx + D f(x)Particular integral F(x) linear polynomial ax + bCx + D exponential functionae kx trig fnacospx or bsinpx or both Ce kx Ccospx + Dsinpx

8. Substituting into the differential equation gives: 4(0) + 4C + Cx + D = 3x + 4 Comparing the coefficients x`s C = 3 no.s 4C + D = 4D = –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 y = Cx + D

9. Ex2Solve Auxiliary equation 2 2 – 1 – 3 = 0 Solving (  0  =  or –1 f(x)Particular integral F(x) linear polynomial ax + bCx + D exponential functionae kx trig fnacospx or bsinpx or both Ce kx where k has the same value as the f(x) term Ccospx + Dsinpx Particular integral is of the form y = Ce 2x. This must satisfy the differential equation Complementary function solutiony = Ae  x + Be –1x

10. Comparing the coefficients e 2x `s8C – 2C – 3C = 1C =    Thus the particular solution is y =    e 2x y = Ce 2x Substituting into the differential equation gives: 2  4Ce 2x – 2Ce 2x – 3Ce 2x = e 2x The complete general solution is given by: y = Complementary function + Particular integral y = Ae  x + Be –1x +    e 2x

11. Solve Complementary function solution y = e 0x (Asin  x + Bcos2x) = Asin2x + Bcos2x Particular integral is of the form y = Ccosx + Dsinx. y = Ccosx + Dsinx f(x)Particular integral F(x) linear polynomial ax + bCx + D exponential functionae kx trig fnacospx or bsinpx or both Ce kx Ccospx + Dsinpx 1 2 + 4 = 0  =  (0 + 2i)as  –4 = 2i Complex roots Here the term is missing so the auxiliary equation becomes

12. y = Ccosx + Dsinx Substituting into the differential equation gives: –Ccosx – Dsinx + 4(Ccosx + Dsinx) = 3sinx + 4cosx 3Ccosx + 3Dsinx = 3sinx + 4cosx Equating coefficients of sinx and cosx 3D = 3D = 1 3C = 4C = The complete general solution is given by: y = Complementary function + Particular integral y = Asin  x + Bcos2x + 1sinx + cosx

13. 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 = x  f(x) as the particular integral. Auxiliary equation 2 2 – 1 – 3 = 0 Solving (2 - 3)( + 1) = 0  =  or–1 Complementary function solution y = Ae  x + Be –1x As f(x) = e –x is contained in the complementary function solution use y = x  Ce –x as the P.I.

14. C.F. y = Ae  x + Be –1x Particular integral is y = x  Ce –x as the f(x) term contains e –x as does the C.F P.I y = x  Ce –x y = Cxe –x Use the product rule to find and Substituting into the differential equation gives:

15. Thus the particular solution is y = –  xe –x The complete general solution is given by: y = Complementary function + Particular integral y = Ae  x + Be –x –  xe –x e –x `s Comparing the coefficients –2C –1C= 1C = – 

16. Complementary function contains f(x) ie RHS of =Particular Integral Ae  x Ce  x Dxe  x AcospxCcospxDxcospx AcoshpxCcoshpxDxcoshpx Rule Multiply the standard Particular Integral by x and then use the product rule to find Summary

17. 1 st order equations which can be solved using the 2 nd order method 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 can be treated as a 2nd order differential equation with Auxiliary equation  – 3 = 0 as    = 3 Complementary function solutiony = Ae 3x Particular integral is of the form y = Ce x

18. P.I. y = Ce x Substituting into the differential equation gives: Ce x – 3Ce x = 2e x –2C = 2 C = –1 The complete general solution is given by: y = Complementary function + Particular integral y = Ae 3x – 1e x

19. Compare this with the method using an integrating factor Solution y = Ae 3x – 1e x Multiply through by e 3x y = Ce 3x – 1e x As was obtained above