(a) Problems based on cauchy-euler

III. (a) problems based on cauchy-euler

 

Example 5.3.1. Solve 2 d²y/dx² – x dy/dx + y = 0 [A.U. April 2002, M/J 2009] [A.U N/D 2002] [A.U Jan. 2016 R15, U.D MA 7151] [A.U A/M 2019 R17]

Solution: Given: [x² D² - xD + 1]y = 0

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

(m − 1)² = 0

m =1, 1

y = (Az + B) e^z

=(A log x + B) x [ x= e^z, z = log x ]

y = = x [A logx + B]

Note: The independent solutions are x, x logx

 

Example 5.3.2. Solve x d²y/dx² +  dy/dx = 0 [A.U. Nov. 2001] [A.U M/J 2005] [A.U Tvli. Jan 2008, Tvli M/J 2009] [A.U N.D 2014 R-13] [A.U N/D 2019, R-17]

Solution: Given: x D²y+ Dy = 0

x² D²y + x Dy = 0 [Multiply by x]

[x² D² + xD] y = 0

Put x =e^z

Convert xy'' + y' = 0 into a linear differential equation with constant coefficients

Ans. upto equation(1)

The auxiliary equation is m²= 0, m= 0, 0

y = [Az + B] e^0z [ since, e^0z = 1]

= Az + B

y = A (logx) + B

 

Example 5.3.3. Solve x²y' + 2xy' + 2y = 0  [AU, May 2001]

Solution: Given: [x² D² + 2xD + 2] y = 0 .... (1)

Put  x = e^z

logx = z

x D = D'

x² D² = D' (D' — 1)

(1) ⇒ [D' (D' - 1) + 2D' + 2] y = 0

[D'₂ - D' + 2D' + 2] y = 0

[D'₂ + D' + 2]y = 0

The auxiliary equation is m² + m + 2 = 0

Example 5.3.4. Solve : x2y'' - 4xy' +6y = x² + logx [A.U N/D 2015 R-13]

Solution: Given (x² D² - 4x D +6)y = x² + log x

Put x = e^z, z = log x

X D = D'

x² D₂ = D' (D' - 1)

The auxiliary equation is m² - 5m + 6 = 0

 (m-2) (m-3) = 0

m=2, m = 3

C.F = Ae^2z + Be^3z

y = C.F + P.I₁+ P.I₂

= Ax² + Bx³ - x² log x + logx6 /6 + 5/36

 

Example 5.3.5. (x²D² + 4x D + 2) y = log x, given that when x= 1, y = 0 and dx/dy = 0.([Note: This is an I.V.P] [A.U. March, 1996][A.U N/D 2013]

Solution: Given: [x² D² + 4xD + 2]y = logx

Put x =e^z

logx = z

xD = D'

x²D² = D' (D' - 1)

[D' (D' - 1) + 4D' + 2]y = z

[D'² - D' + 4D' + 2]y = z

[D'² + 3D' + 2]y = z

The auxiliary equation is m² + 3m + 2 = 0

(m + 2) (m + 1) = 0

 (i) Given: x = 1, y (1) = 0

 (1) ⇒ y (1) = A + B + [2 log 1 - 3]

0 = A + B + 1 [0 − 3] [ log 1 = 0]

0 = A + B – 3/4

⇒ A+B .....(2)

(ii) Given: x = 1, y' (1) = 0

0 = - 2A – B + 1/2

2A + B = ½  ......(3)

(2) – (3 ) ⇒ -A = ¾ - ½ = 1/4

A = -1/4

(2) ⇒ -1/4 + B = 3/4

⇒ B = ¾ + ¼ = 4/4 =1

 

Example 5.3.6. Solve the equation (x²D² + 3xD+5) y = x cos (log x) [A.U M/J 2009]

Solution: Given: (x² D² + 3x D + 5) y = x cos (log x).....(1)

Put x =e^z

log x = z⇒ z = log x

xD = D' where D' = d/dz

x² D² = D' (D' -1)

(1) ⇒ [D' (D' - 1) + 3D' + 5] y = e^z cos (z)

[D'² - D' + 3D'+5]y = e^z cos z

 [D'² + 2D' + 5]y = e^z cos z

The auxiliary equation is m² + 2m + 5 = 0

i.e., P.I = 7/65 x cos (log x) + 4/65 x sin (log x)

 

Example 5.3.7. Solve [x² D² - xD + 1] y = [log x/x]² [A.U April 02, N/D 07] [A.U M/J 2014] [A.U N/D 2018 R-17]

Solution: Given: [x² D² - xD + 1] y = [log x/x]²

Put x = e^z

log x = z

xD = D'

x² D² = D' (D' − 1)

 [D' (D' - 1) - D' + 1] y = [z/e^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 + B)x

 

Example 5.3.8. Solve (x²D² - 2xD - 4) y = x² + 2log x[AU, May 1998] [A.U M/J 2010]

Solution: Given: (x²D² - 2xD - 4) y = x² + 2 logx .....(1)

Put x = e^z

log x = z

x D = D'

x² D2 = D' (D' - 1)

 [D'^z - 3D' - 4]y = e^2z + 2z

The auxiliary equation is m² - 3m - 4 = 0

m = 4, m = -1

y =A/x + Bx⁴ – 1/6x² – log x + 3/8

 

Example 5.3.9. Solve d²y/dx² + 1/x dy/dx = 12 log x/x² [ A.U N/D 2012]

Solution: Given: d²y/dx² + 1/x dy/dx = 12 log x/x²

i.e., x² d²y/dx² + x dy/dx = 12 log x

i.e.,[ x²D² + xD] = 12 log x ......(1)

put x = e^z

log x = z

xD =  D'

x² D² = D' (D' - 1)

[D' (D' - 1) + D']y = 12z

[D'² - D' + D']y = 12z

(D')²y = 12z

The auxiliary equation is m² = 0

C.F = (Az + B) e^0z = Az + B = A logx + B

= 2z³ = (logx)³

y = C.F+ P.I

y = A log x + B + 2 (logx)³

 

Example 5.3.10. Solve (x² D² + 4xD + 2) y = x log x [A.U M/J 2016 R-8]

Solution: Given: (x² D² + 4xD + 2) y =x log x .....(1)

put x =e^z

z = log x

xD = D'

x² D² = D' (D' - 1)

 [D' (D' - 1) + 4D' + 2]y = e^z z

[D'² + 3D' + 2]y = ze^z

The auxiliary equation is m² + 3m + 2 = 0

⇒ m =-2, m = -1

C.F = Ae^-2z + Be^-z

= 1/36 x (6 log x - 5)

 y = C.F + P.I

(i.e.,) y = A (1/x²) + B (1/x) + 1/36 x (6 logx - 5)

 

Example 5.3.11. Solve : (x² D² - xD + 1) y = sin (log x)[A.U N/D 2014 R-13]

Solution: Given: (x² D² - xD + 1) y = sin (log x)

Put x = e^z

Z = sin (log x)

xD= D'

x² D² = D' (D' - 1), where D' = d/dz

(1) ⇒ (D' (D' - 1) - D' + 1) y = sin z

The auxiliary equation is

(m − 1)² = 0

i.e., m = 1, 1

C.F = (Az + B) e^z

The general solution is

y = C.F + P.I

= (A logx + B)x + ½ cos (logx)

 

Example 5.3.12. Solve : x² d²y/dx² – x dy/dx + y = log x + π [A.U A/M 2015 R13] [A.U N/D 2016 R-13]

 Solution: Given: [x²D2 - xD + 1]y =log x + π

Put x = e^z

Log x = z

So that xD = D'

x² D² = D' (D' — 1)

= [D' (D' - 1) - D' + 1]y = z + π

The auxiliary equation is (m-1)² = 0

m = 1,1

C.F = (Az + B)e^z = (A logx + B) x

y = C.F + P.I

y = (A log x + B)x + log x + л + 2

 

Example 5.3.13. Solve the differential equation (x² D² - xD + 4) y = x² sin (log x) [A.U N/D 2009, M/J 2012]

Solution : Given: (x² D² - xD + 4) y = x² sin (log x).....(1)

Put x = e^z

z = log x

xD = D'

x² D² = D' (D' − 1)

(1) ⇒ [D' (D' − 1) − D' + 4]y = (e^z)² sin (z)

[D'² - D' - D' + 4 ]y = e^2z sin z

[D'² - 2D' + 4]y = e^2z sin z

The auxiliary equation is m² - 2m + 4 = 0

m² - 2m + 4=0

 

Example 5.3.14. Solve x²d²y/dx² + 4x dy/dx = e^x [ A.U Tvli. A/M 2009 ]

Solution: Given: (x² D² + 4x D + 2) y = e^x

Put x = e^z   ⇒ logx = z

So that,

x D = D'

x² D² = D' (D' - 1)

The auxiliary equation is (m + 2) (m + 1) = 0 ⇒ m = -2, m = -1

y = C.F + P.I

y = C₁1/x + C₂ 1/x² + 1/x² e^x

 

Example 5.3.15. Solve x³d³y/dx³ + 2x² d²ydx² + 2y = 10 [ x + 1/x ]

Solution : Given: [x³ D³ + 2x² D² + 2]y = 10 [x + 1/x ]

Put x = e^z   ⇒ logx = z

So that, x D = D' ⇒x² D² = D' (D' - 1) ⇒x³ D³ = D' (D' - 1) (D' - 2)

 [D' (D' - 1) (D' - 2) + 2D' (D' - 1) + 2] y = 10 [e^z + e^-z]

The auxiliary equation is m³ - m² + 2 = 0

m = -1, m=1 ± i

 C.F = C₁e^-z + e^z [C₂ cos z + C₃ sin z]=C1 (e^z)^-1 + e^z [C₂ cos z + C₃ sin z]

y = C.F+ P.I₁ + P.I₂

y = C₁ 1/x + [C₂ cos (log x) + C₃ sin (log x)]x + 5x + 2 log x/x

 

Example 5.3.16. Solve x²d²y/dx² + x dy/dx + y = log x sin (log x)

Solution: Given: [x² + D² + xD + 1]y = log x sin (log x)

Put x = e^z     ⇒z = log x

so that, xD = D' , x² D² = D' (D' − 1)

[D' (D' − 1) − D' + 1]y = z sin z

[D'² - D' - D' + 1 ]y = z sin z

[D'² + 1 ]y = z sin z

The auxiliary equation is m² + 1= 0 ⇒m = ±i

P.I = − (logx)² cos (log x)/4 +  (logx) sin (log x)/4

y = C.F + P.I

 

Example 5.3.17. Solve x³d³y/dx³ + 3x²d²y/dx² + x dy/dx + 8y = 65 cos (log x)

Solution: Given: [x³ D³ + 3x² D² + D + 8]y = 65 cos (log x)

Put x = e^z , z = log x

so that, xD = D' , x² D² = D' (D' − 1)

x³ D³ = D' (D' − 1) (D' − 2)

 

Example 5.3.18. Solve : x²d²y/dx² + 2x dy/dx -12y = x³ log x

Solution: Given: [x² D² + 2xD - 12]y =x³ log x

Put x = e^z

z = log x

xD = D'

x² D² = D' (D' − 1)

y = C.F + P.I

y = C₁x^-4 + C₂x³ + x³/98 log x [7 logx - 2]

 

Example 5.3.19. Solve x²d²y/dx² - 3x dy/dx + 4y = (1 + x)²

Solution: Given: (x² D² - 3xD + 4) y = (1+x)² .......................(1)

Put x = e^z

z = log x

xD = D'

x² D² = D' (D' − 1)

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

The auxiliary equation is (m – 2)²= 0 , m = 2,2

CF = (C₁z + C₂) e^2z

P.I₃ = 2e^z =2x

y = C.F + P.I₁ + P.I₂ + P.I₃

y = (C₁ logx + C₂) x² + 1/4 + x² (log x)²/2 + 2x

 

Example 5.3.20. Solve [x²D² - (2m - 1) x D + (m² + n²)ly = n² x^m log x

Solution: Given: [x²D² - (2m - 1)xD + (m² + n²)ly = n²x^m logx

Put x = e^{z   ,} z = log x = z

xD = D' , x² D² = D' (D' − 1)

Example 5.3.21. Reduce x²d²y/dx² - 3x dy/dx + 3y = x  into a differential equation with constant coefficients. [A.U. N/D 2007]

Solution : Given: x²d²y/dx² - 3x dy/dx + 3y = x

i.e., [x² D² - 3xD + 3]y = x ... (1)

Put x = e^z

 log x = z

xD = D' , x² D² = D' (D' − 1)

(1) ⇒ [D' (D' − 1) − 3D' + 3]y = e^z

 [D'² - D' - 3D' + 3 ]y = e^z

[D'² - 4D' + 3]y =e^z

 

Example 5.3.22. Write Euler's Homogeneous linear differential equation. How will you convert it to a linear differential equation with constant coefficients? [A.U. A/M. 2008]

Solution: An equation of the forma

where a₀, a₁, a₂,+...+a_n..., an are constants and f(x) is a function of x.

Equation (1) can be reduced to linear differential equation with constant coefficients by putting the substitution.

 x = e^z  (or) z = log x xD = D' , D' = d/dz

x² D² = D' (D' − 1), x³ D³ = D' (D' − 1) (D' − 2)

 

Example 5.3.23. TrAnsform (x² D² + xD + 1) y = 0 into differential equation with constant coefficients, where D = d/dx [A.U. CBT M/J 2008 SM0101]

Solution: Given: (x² D² + xD + 1) y = 0 .....(1)

Put x = e_z   log x = z

 xD = D' ; x² D² = D' (D' − 1)

(1) ⇒ [D' (D' − 1) + D' + 1]y = 0

[D'2 - D' + D' + 1 ]y = 0 ,[D'² + 1]y = 0

 

5.3.23 (b) TrAnsform the equation xy" + y' + 10 into a linear equation with constant co-efficients. [A.U N/D 2018 R-17]

Solution: Given: xy" + y' + 1 = 0

 x2y" + xy' + x = 0

x2y" + xy' = -x ......(1)

Put x = e^z  log x = z , xD = D', x² D² = D' (D' − 1)