5. (a) D² + 2pD + (p² + q²)y = 0

It's auxiliary equation is

m² + 2pm + p² + q² = 0

⇒ (m + p)² + q² = 0

⇒ m + p = \(\pm q\)i

⇒ m = -p \(\pm q\)i

\(\therefore\) C.F. = e^-px (C₁ cos9x + C₂sin qx)

Hence, y = e^-px (C₁cos 9x + C₂ sin 9x) is solution of given differential equation.

(b) m² + 1 = 0

⇒ m = \(\pm i\)

C.F. = C₁ cos x + C₂ sin x

y₁ = cos x, y₂ = sin x

w(y₁, y₂) = \(\begin{vmatrix}cosx&sin x\\-sin x&cos x\end{vmatrix}\) = cos²x + sin²x = 1

u₁ = \(\int\frac{\begin{vmatrix}0&sin x\\cosxsin3x&cosx\end{vmatrix}}{W(y_1,y_2)}dx\) = \(\int-sin xcos xsin 3xdx\)

 = \(\frac{-1}2\int sin2xsin 3x dx\) 

 = \(\frac{-1}4\int 2sin2xsin 3x dx\) 

= \(\frac{-1}4\int4[cos(2x-3x)-cos(2x+3x))dx\)

 = \(\frac{-1}4[\int cos xdx-\int cos 5xdx)\) 

 = \(\frac{-1}4[sin x - \frac{sin 5x}5]\)

u₂ = \(\int\frac{\begin{vmatrix}cos x&0\\-sin x&cosxsin 3x\end{vmatrix}}{1}dx\) 

 = \(\int\)cos²x sin 3x dx

 = \(\int\) \(\frac{1+cos2x}2sin3xdx\)

 = \(\frac12\)\(\int\)sin 3x dx + \(\frac14\)\(\int\)2sin 3x cos 2x dx

 = -\(\frac12\) \(\frac{cos 3x}3+\frac14\int\)(sin(3x + 2x) - sin (3x - 2x)]dx

 = -\(\frac16\)cos3x + \(\frac14\int\) sin 5x dx - \(\frac14\int\) sin x dx

 = -\(\frac16\) cos 3x - \(\frac1{20}\) cos 5x + \(\frac14\) cos x

\(\therefore\) P.I. = u₁y₁ + u₂y₂ 

  = -\(\frac14\)sin x cos x + \(\frac1{20}\) cos x sin 5x - \(\frac1{6}\)sin x cos 3x

- \(\frac1{20}\) sin x cos 5x + \(\frac1{4}\) sin x cos x

 = \(\frac1{20}\) (cos x sin 5x - sin x cos 5x) - \(\frac1{6}\) sin x cos 3x

\(\therefore\) y = C.F. + P.I.

= C₁ cos x + C₂ sin x + \(\frac1{20}\) (cos x sin 5x - sin x cos 5x)

- \(\frac1{6}\)sin x cos 3x

(c) m³ + m² + m = 0

m(m² + m + 1) = 0

m = 0, \(\frac{-1\pm\sqrt3i}2\) 

C.F. = C₁ + e^-x/2(C₂ cos(\(\frac{\sqrt{3x}}{2}\)) + C₃ sin(\(\frac{\sqrt{3x}}{2}\)))

P. I. = \(\frac{1}{D^3+D^2+D}\) (e^2x + x² + x)

 = \(\frac{e^{2x}}{8+4+2}+\frac1D\)(1 + (D + D²))^-1 (x² + x)

 = \(\frac1{14}e^{2x}\)+ (\(\frac1D\) - 1 - D + D + 2D² - D²) (x² + x)

 = \(\frac1{14}e^{2x}\) + (\(\frac1D\) - 1 + D²) (x² + x)

 = \(\frac1{14}e^{2x}\) + \(\frac{x^3}3-\frac12x^2-x+2\)

y = C.F. + P.I.

 = C₁ + e^-x/2(C₂cos(\(\frac{\sqrt{3x}}{2}\)) + C₃sin(\(\frac{\sqrt{3x}}{2}\)))

 + \(\frac1{14}e^{2x}\) + \(\frac{x^3}3-\frac{x^2}2-x+2\) 

6. (a) Let x = e^z

Then xd = \(\frac d{dz}=D'\)

x²D² = D' (D' - 1)

Given D.E. converts to

(D' (D' - 1) - 4D' + 6)y = e^yz

⇒ ((D')² - 5D' + 6)y = e^yz

m² - 5m + 6 = 0 by taking D' = m

⇒ m = 2, 3

C.F. = C₁e^2z + C₂e^3z

P.I. = \(\frac{1}{(D')^2 - 5D' + 6}e^{yz}\)

 = \(\frac{e^{4z}}{4^2-5\times4+6}\) = \(\frac{e^{4z}}2\)

\(\therefore\) y = C.F. + P. I.

 = C₁(e^z)² + C₂(e^z)³ + \(\frac12\)(e^z)⁴

 = C₁x² + C₂x³ + \(\frac{x^4}2\) (\(\because\) e^z = x)

(b) (x²D² + xD + 1)y = log x + cos (log x)

Let x = e^z

Then  given D.E. converts to

(D'(D'- 1) + D' + 1)y = a + cos z

⇒ ((D')² + 1)y = z + cos z

\(\therefore\) m² + 1 = 0

m = \(\pm i\)

C.F. = C₁cos z + C₂ sin z

P.I = \(\frac1{(D')^2+1}\)(z + cos z)

 = (1 - (D')² + (D')⁴+....) z + \(\frac{z}2\)sin z

 = z + \(\frac{z}2\) sin z

\(\therefore\) y = C.F. + P. I.

= C₁cos z + C₂ sin z + z + \(\frac{z}2\) sin z

= C₁cos(log x) + C₂ sin (log x) + log x + \(\frac{log x}2\) sin (log x)

(c) (x + 1)²\(\frac{d^2y}{dx^2}\) - 3(x + 1) \(\frac{dy}{dx}\) + 4y = x

Let x + 1 = e^z

Then D.E. converts to

(D'(D'-1) - 3 D' + 4)y = e^z - 1

⇒ ((D')² - 4D' + 4)y = e^z - 1

⇒ (D' - 2)²y = e^z - 1

\(\therefore\) (m - 2)² = 0

m = 2, 2

C.F. = (C₁ + C₂z)e²z

P.I. = \(\frac{1}{(D'-2)^2}(e^z-e^{0z})\)

 = \(\frac{e^z}{(1-2)^2}-\frac{e^{0z}}{(0-z)^2}\)

 = \(\frac{e^z}1-\frac14\)

\(\therefore\) y = C.F. + P. I.

= (C₁ + C₂z)e²z + e^z - \(\frac14\)

 = (C₁ + C₂ log x) x² + x - 1/4 is solution of given differential equation.