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By Heena Nainwani
(13026017017)
2nd Order Linear homogenous diff.
equation with constant co-efficient.
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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)
Method of finding particular Integral
Y=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
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For example:
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Q: (D2-5D+6)y= e3x
PI=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
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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)
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For example:
Q: (D2-2D+5)y= sin3x
 PI =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)
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CASE3: To find PI when R(x) is of the form xm
or polynomial of degree of m where m Is
positive integer
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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.
SOME EXPANSIONS……
[1- X]-1 =1+X+X2+X3+……
 [1+X]-1 =1-X+X2-X3+……
 [1- X]-2 =1+2X+3X2+4X3+……
 [1+X]-2 =1-2X+3X2 -4X3+……
 [1- X]-3 =1+3X+6X2+10X3+……
 [1+X]-3 =1-2X+6X2 -10X3+……
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For example:
Q: (D3 + 8) y = x4 + 2x +1
 PI =1/(D3+8)*(x4+2x+1)
=1/8(1+D3/8)*(x4+2x+1)
=1/8(1+D3/8)-1(x4+2x+1)
=1/8[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]
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CASE4: To find PI when R(x) is of the form
eax *v where v is any function of x.
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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]
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Method of Undeterminant coefficient:
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
For example:
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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-9Bcos3x4[Asin3x+Bcos3x]=2sin3x
=-5Asin3x-5Bcos3x=2sin3x
A=(-2/5) and B=0
Yp =-2/5sin3x
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’
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For example:
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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
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]
Cauchy Euler Equation
A differential equation of the form
(Xn dny/dxn )+a1x(n-1) d(n-1)y/dxn1+a …….+tanyx-------2
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
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Lengendre’s Linear Equation
(a1x+b1)n dny/dxn +a1(a2x+b2)(n-1) d(n1)y/dxn- 1+a …….+a yx-------2
n
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
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THANKS……