1 HIGHER ORDER By Heena Nainwani ( )

2 2nd Order Linear homogenous diff. equation with constant co-efficient.The homogeneous equation is y’’+ay’+by=0 consider y=emx then y’=emx m ,y’’=emx m2 Then the equation formed is m2+am+b=0 Above equation is known as auxiliary equation. Case1: If roots are distinct then y=c1em1x +c2em2x Case2: : If roots are equal y=(c1 +c2 x) emx Case 3: If roots are complex then y=epx (cosqx+isinqx)

3 Method of finding particular IntegralY=CF+PI To find PI when R(x) is of the form eax where a is a constant. Result: PI = 1/f(D) eax = 1/f(D) eax ,f(a)!=0 = x*1/f(D) eax ,f(a)=0

4 For example: Q: (D2-5D+6)y= e3xPI=1/(D2-5D+6) * e3(Here instead of D put 3 ) =x*1/(2D-5)* e3x =x*1/2* e3x =x1/2 e3x CI=m2-5D+6 =(m-3)(m-2) m=3 and m=2. Here roots are equal y=C.F+P.I = c1e3x +c2e2x +x*1/2 e3x

5 CASE2: To find PI when R(x) is of the form sinax or cosax.RESULT: 1/P(D)2 sinax=1/Ø{-a2}*sinaxØ{-a2}≠ 0 = X 1/Ø’{-x2}≠ 0 sinax Ø(-a2 )=0 Where f(D) = Ø(D2)

6 For example: Q: (D2-2D+5)y= sin3xPI =1/(D2-2D+5)*sin3x(Put the value of only in D2 ) =1/-9-2D+5*sin3x =1/-4-2D*sin3x =-1/2*(1/(D+2)*sin3x) =-1/2*((D-2)/(D2-4)*sin3x) =-1/2((D-2)/-9-4)*sin3x) =1/26*((D-2) sin3x) =1/26(Dsin3x-2sin3x) =1/26(3cos3x-2sin3x)

7 CASE3: To find PI when R(x) is of the form xm or polynomial of degree of m where m Is positive integer RESULT: R(x)= xm then expand [1±Ø(D)]-1 upto the term containing Dm only because term after Dm operation upon xm become zero.

8 SOME EXPANSIONS…… [1- X]-1 =1+X+X2+X3+…… [1+X]-1 =1-X+X2-X3+……

9 For example: Q: (D3 + 8) y = x4 + 2x +1 PI =1/(D3+8)*(x4+2x+1)=1/[x4+2x+1-1/8(D3{x4+2x+1}] =1/8[x4+2x+1-1/8(24x)] =1/8[x4-x+1]

10 CASE4: To find PI when R(x) is of the form eaxCASE4: To find PI when R(x) is of the form eax *v where v is any function of x. RESULT: 1/f(D)* eax *v = eax *1/f(D+a)*v For example: Q: (D2-2D+1)y= e3x * x2 PI =1/(D2-2D+1)* e3x * x2 (instead of D we put (D+3) = e3x *1/(D+3)2-2(D+3)+1)*x2 = e3x *1/(D2+6D+9-2D-6+1)*x2 = e3x * 1/D2+4D+4*x2 = e3x [1/4(1+(D+D2/4)]*x2 = e3x /4 [1+(D+D2/4)-1*x2 = e3x /4[1-(D+D2/4) +(D+D2/4)2]*x2 = e3x /4 [x2-(2x+2/4)+2] = e3x/4 [x2-2x+3/2]

11 Method of Undeterminant co-efficient:To find PI as following tables: RHS of f(D)y=x Trial solution X=eax Yp=Aeax X=sinax/cosax Yp= Asinax+Bcosax X=c Yp=A X=ax+b Yp=A+Bx X=ax3+bx2+cx+d Yp=A+Bx+Cx2+Dx3 X=ax2+bx/ax2+b Yp=A+Bx+Cx2 X=x*eax Yp=eax(A+Bx) x=x2eax Yp=eax(A+Bx+cx2) x=asinax Yp=sinax(A+Bx)+ cosax(A+Bx) x=e2x Yp=Ae2x x=e2x-3e-x Yp=Ae2x+Be-x

12 For example: Q: (D2+4)y=2sin3x Here to find Yp use above table:So, Yp=Asin3x+Bcos3x (as comparing of 2sin3x) Y’p= 3Acos3x-B3sin3x Y’’p=-A*9sin3x-9Bcos3x Y’’+4y=2sin3x =-A9sin3x-9Bcos3x- 4[Asin3x+Bcos3x]=2sin3x =-5Asin3x-5Bcos3x=2sin3x A=(-2/5) and B=0 Yp =-2/5sin3x

13 Method of variation of parameter:Yp=-y1ʃy2R(x)/w dx +y2ʃy1(-R(x))/w where, y1 and y2 are: Y’’+P(x) y’ + d(x)y=0 And w=y1y2’ - y2y1’

14 For example: Q: Using method of variation of parameter for(D2-6D+9)y=e3x/x2 A.E is m2-6m+9=0 (m-3)2=0 m=3,m=3 Yc=(c1+c2x)e3x =c1e3x+c2xe3x Let y1 = e3x , y2 = xe3x

15 W= y1y2’ - y2y1’ =e6x Putting values of y1, y2 and w and R in standard equation Yp=-y1ʃy2R(x)/w dx +y2ʃy1(-R(x))/w =-e3xʃxe3x*e3x/x2e6x *dx+xe3xe3x/x2e6x* dx =-e3xʃ1/x dx+x e3x-ʃx-2dx =-e3x[logx+1]

16 Cauchy Euler Equation A differential equation of the form(Xn dny/dxn )+a1x(n-1) d(n-1)y/dxn- 1+a2…….+tanyx Where a1,a2,…..an are constant and x is a function of n is called cauchy euler equation. Method:Put x=e2 xdy/dx=Dy x2 d2y/dx2=D(D-1)y x3 d3y/dx3=D(D-1)(D-2)y

17 Lengendre’s Linear Equation(a1x+b1)n dny/dxn +a1(a2x+b2)(n-1) d(n- 1)y/dxn- 1+a2…….+anyx where a1,b1,a2,b2 are constant and x is a function of x is called lengendre’s linear equation. Method Put ax+b=e2 (ax+b)dy/dx= a Dy (ax+b)2d2y/dx2=a2D(D-1)y (ax+b)3d3y/dx3=a3D(D-1)(D-2)y

18 THANKS……