(b) problems based on legendre's linear differential equation

III. (b) PROBLEMS BASED ON LEGENDRE'S LINEAR DIFFERENTIAL EQUATION

(EQUATION REDUCIBLE TO LINEAR FORM)

An equation of the form

where k's are constants and Q is a function of x is called Legendre's linear differential equations. Such equations can be reduced to linear equations with constant coefficients by putting

ax + b = e^z

 (i.e.,) z = log (ax + b )

If D' = d/dz , then (ax + b)D = aD'

 (ax+b)² D² = a² D' (D' - 1) and so on.

After making these substitutions in (1), it reduces to a linear differential equation with constant coefficients.

Legendre's linear differential equation can be reduced to the Euler homogeneous linear forms also, by putting ax + b = t and then solved by the Euler's method.

Example 5.3.24. Solve (x + 2)²d²y/dx² – (x+2)dy/dx + y = 3x + 4  [A.U. Nov. 2001] [A.U A/M 2019 R-17] [A.U. N/D 2019, R-17]

Solution: Given: [(x+2)² D² - (x + 2) D + 1]y =3x + 4 .....(1)

Put x + 2 = e^z ⇒ x = e^z - 2

log (x+2)= z

(x+2) D = D'

(x + 2)² D² = D' (D' - 1)

[D' (D' — 1) — D' + 1]y = 3 (e^z - 2) + 4

[D'²-D' - D' + 1]y = 3e^z -6 + 4

[D'² - 2D' + 1]y = 3e^z - 2

The auxiliary equation is m² - 2m + 1 = 0

(m − 1)² = 0 ⇒ m = 1, 1

C.F = (Az + B) e^z =[A log (x + 2) + B] (x + 2)

y = C.F+ P. I₁ + P. I₂

y = [A log (x + 2) + B] (x + 2) + 3/2 [log (x + 2)]² (x + 2) − 2

Example 5.3.25. Solve [(x + 1)² D² + (x + 1) D + 1] y = 4 cos [log (x + 1)][A.U. Nov. 2001] [A.U N/D 2011 R-8] [A.U Jan. 2018 R-17]

Solution: Given: [(x + 1)² D² + (x + 1) D + 1]y = 4 cos [log (x + 1)]

Put 1 + x =e^z

z = log (1 + x)

(x+1)D = D'

(x + 1)² D² = D' (D' -1)

 [D' (D' + D' + 1]y = 4 [cos z]

[D'² - D' + D' + 1]y = 4 cos z

 [D'² + 1]y = 4 cos z

The auxiliary equation is m² + 1 = 0

m = ± i

C.F. = [A cos z + B sin z]

= A cos [log (x + 1)] + B sin [log (x + 1)]

= 2 log (1 + x) sin [log (1 + x)]

y =  C.F. + P.I.

y = A cos [log (x + 1)] + B sin [log (x + 1)] + 2 log (x + 1) sin [log (x + 1)]

Example 5.3.26. Solve (1+x)² d²y/dx² + (1 + x)dy/dx + y = 2 sin [log (1 + x)]  [A.U A/M 2011 R-8, A/M 2017 R-13]

Solution: Given: [(1 + x)² D² + (1+x) D + 1]y = 2 sin [ log (1 + x)

Put 1+ x= e^z

z = log (1 + x)

(x + 1) D = D'

(x+1)² D² = D' (D' - 1)

[D' (D' - 1) + D' + 1]y = 2 sin z

y = C.F + P.I

y = A cos [log (1+x)] + B sin [log (1 + x)] - log (1+ x) cos [log (1+x)]

Example 5.3.27. Solve (3x+2)² D² + 3 (3x + 2)D - 36 = 3x² + 4x + 1 [A.U M/J 2013 R-8]

Solution: Given: [(3x+2)² D² + 3 (3x+2) D - 36] y = 3x² + 4x + 1

Put 3x + 2 = e^z

log (3x + 2) = Z

3x = e^z – 2

x =1/3e^z – 2/3

Let

Example 5.3.28. Solve (2x+3)² d²y/dx² - (2x+3)dy/dx - 12y = 6x [A.U N/D 2002, Jan. 2005]

Solution: Given: (2x+3)² d²y/dx² - (2x+3)dy/dx - 12y = 6x

i.e., [(2x+3)² D² - (2x + 3) D - 12]y = 6x

Put 2x + 3 = e^z

log (2x+3)= Z

2x = e^z - 3

x = e^z/2 – 3/2

Let (2x+3) D = 2D'

 (2x+3)² D² = 2² D' (D'-1) = 4D' (D'-1))

Example 5.3.29 TrAnsform the equation (2x+3)² d²y/dx2 - 2(2x+3)dy/dx - 12 y = 6x into a linear differential equation with constant coefficients. [A.U M/J 2012] [A.U A/M 2015 R-13]

Solution : Given: (2x+3)² d²y/dx2 - 2(2x+3)dy/dx - 12 y = 6x dx2

i.e., [(2x+3)² D² - 2 (2x+3) D - 12] y = 6x ....(1)

Put 2x + 3 = e^z

log (2x+3)= z

2x = e^z – 3 ,  x = e^z/2 – 3/2

Let

(2x+3) D = 2 D'

(2x+3)² D² = 2² D' (D' - 1) = 4D' (D' -1)