1) The document discusses Legendre's linear differential equation with variable coefficients. It provides the general form of such an equation and defines operators to simplify it. 2) It then works through solving a specific example equation by making a change of variables and rewriting the differential operators. 3) The solution obtained involves exponential functions of the new variable along with arbitrary constants.

1. sachin patel
130940109087

2. Legendre's eqaution

3.  Legendre's linear diffrential equation with variable
coeffcients:-
legendre's linera diffrential eqaution variable coeffcients of an order is of
the form:-
(Ax+b)n .dny/dxn +k1(ax+b)n-1. dn-1y/dxn-1 +
k2(ax+b)n-2.dn-2y/dxn-2 +.......kny=R(x)
& Assume that:-
(ax+b) d/dx=a.D
(ax+b)2.d2/dx2=a2.D(D-1)
(ax+b)3.d3/dx3=a3.D.(D-1).(D-2)

4. & we condsider...
log(ax+b)=z
ax+b=ez
solve ; (3x+2)2d2y/dx2 + 3(3x+2)dy/dx-36y=3x2+4x+1.......(1)
Ans. log(3x+2)=z
3x+2=ez
3x=ez-2
x=ez-2/3
(3x+2)2.d2/dx2=9D.(D-1).............(2)
(3X+2).dy/dx=3D .............(3)
eqn (2 )& (3)..using a .....eqn (1)

5.  9D.(D-1)y + 3(3D)y-36y=3{ez-2/3} +4{ez-2/3}+1
 y[9D2-9D+9D-36]=3{(e2z-4ez+4)/9}+4{(ez-2)/3}+1
 [9D2-36]y=(e2z-4ez+4+4ez-8+3)/3
 [D2-4]y=1/27.(e2z-1)
f(x)y=R(x)
axuallary eqn= (m2-4)=0
m2=2
m=±2

6. ycf=c1e2z+c2e(-2z)
 c1(ez)2+c2(ez)-2
ycf=c1(3x+2)2+c2(3x+2)-2
ypI= 1/27(D2-4).[e2z-1]
 1/27.{e2z/(D2-4)-1/(D2-4)}
 1/27{z.e2z/2D - (-1/4)}

7.  1/27[z.e2z/4 + 1/4]
 1/27[log(3x+2)/4.(3x+2)2+1/4]
Ans. 1/108[(3x+2)2log(3x+2) + 1]

8. Ex- x4y'''+2x3y''-x2y'+xy=1
x3y'''+2x2y"-xy'+y=1
assume ..logx =z
[D(D-1)(D-2)y]+2[D(D-1)y]-Dy+y=1/x
[D3-3D2+2D]y+[2D2-2D]y-Dy+y=1/X
y[D3-3D2+2D+2D2-2D-D+1]=1/X
y[D3-D2-D+1]=1/X
Ans

9. axuallary eq:-
m3-m2-m+1=0
m2(m-1)-(m-1)=0
(m-1)(m2-1)=0
m=1 , m=±1
ycf=(c1+c2z)ez+c3e-z
(c1+c2logx)x+c3x-1
ypI= e-z/D3-D2-D+1
D→a=-1 so ypi=0

10. so,
z.e-z/3D2-2D-1
z.e-z/3+2-1
z.e-z/4
ypI=(logx.x-1)/4

11. Thank You
much mor fun....... sachinpatel1100@gmail.
com