# M III CS/IT May 2016

Q.1 a i) Solve ${(D^2-D)y = e^x sin (x)}$

Solution : The auxiliary equation is ${m^2-m =0}$ or ${m(m-1)=0}$. The roots are ${m_1 = 0}$ and ${m_2 = 1}$. The complimentary function will be

${y_c = c_1 e^{0x} + c_2 e^{1x} = c_1 + c_2 e^x}$

The particular integral will be

${y_p = \frac {1}{D^2-D} e^x \cdot sin (x)}$

Using ${\frac {1}{\phi (D)} e^{ax} V = e^{ax} \times \frac {1}{\phi (D+a)} V}$,

${y_p = e^x \times \frac {1}{(D+1)^2 - (D+1)} sin (x)}$

${= e^x \times \frac {1}{D^2+2D+1-D-1} sin (x) = e^{x} \times \frac {1}{D^2+D} sin (x)}$

Using ${\frac {1}{\phi (D^2)} sin (ax) = \frac {1}{\phi (-a^2)} sin (x)}$,

${y_p = \frac {1}{-1+D} sin(x) = e^x \frac {1+D}{[-1+D][1+D]} = e^x \times \frac {1+D}{D^2-1} sin(x) = e^x \times \frac {(1+D)}{-2} sin(x)}$

Therefore,

${e^x \times \frac {-1}{2} \left (sin (x) + \frac {d}{dx} sin(x) \right ) = \frac {-e^x}{2} [sin (x) + cos (x)]}$

The complete solution is

${y =y_c + y_p = c_1 + c_2 e^x - \frac {e^x}{2} [sin (x) + cos (x)]}$

Q.1 a ii) Solve ${\frac {dx}{3z-4y} = \frac {dy}{4x-2z} = \frac {dz}{2y-3x}}$

Solution : These are symmetric simultaneous differential equations. We will find 2 sets of values ${l,m,n}$ which will give ${ldx + mdy +ndz = 0}$.

Note the pattern in the ratios. In the first ratio, ${2}$ and ${x}$ are absent in the denominator, in the second, ${3}$ and ${y}$ and in the third ${4}$ and ${z}$.

If we choose the first set of multipliers as ${2,3}$ and ${4}$, we get,

${2(3z-4y) + 3 (4x-2z) + 4 (2y-3x)}$

${= 6z-8y+12x-6z+8y-12x = 0}$

Hence, ${2dx+3dy+4dz = 0}$

On integrating,

${\int2 dx + \int 3 dy + \int 4 dz = 2x+3y+4z = c_1}$

Note that this represents a family of planes normal to the position vector of the point ${(2,3,4)}$.

If we choose the second set of multipliers as ${x,y}$ and ${z}$, we get

${x(3z-4y) + y(4x-2z) + z (2y-3x)}$

${= 3zx-4xy+4xy-2yz+2yz-3zx = 0}$

Hence, ${xdx + ydy + zdz = 0}$

On integrating,

${\int x dx + \int y dy + \int zdz = \frac {x^2}{2} + \frac {y^2}{2} + \frac {z^2}{2} = c_2}$

Note that this represents a family of spheres with center at origin.

Q.1 a iii) Solve ${(D^2+9)y = x^2 +2x + cos (x)}$

Solution : The auxiliary equation is ${m^2+9 =0}$. Its roots are ${0+3i}$ and ${0-3i}$. These are complex conjugate of each other. Hence, the complimentary function will be

${y_c = e^{0x}[c_1 cos (3x) + c_2 sin (3x)] = c_1 cos (3x) + c_2 sin (3x)}$

The particular integral will be

${\frac {1}{D^2+9} x^2 + x + cos (3x) = \frac {1}{D^2+9} x^2 + \frac {1}{D^2+9} 2x + \frac {1}{D^2+9} cos (3x)}$

Note that first two functions on RHS are algebraic functions with integral powers of ${x}$. Hence, we need to use the binomial expansion in the form ${\frac {1}{1+ \phi (D)}}$.

The third function is ${cos (3x)}$. If we use ${\frac {1}{\phi (D^2)} cos (ax) = \frac {1}{\phi (-a^2)} cos (ax)}$, we get ${\frac {1}{0}}$, so we need to use the alternative, which is ${\frac {1}{D^2+a^2} cos (ax) = \frac {x}{2a} sin (ax)}$.

Considering all these,

${y_p = \frac {1}{9} \left [\frac {1}{\frac {D^2}{9} + 1} \right ] x^2 + \frac {1}{9} \left [\frac {1}{\frac {D^2}{9} + 1} \right ] 2x + \frac {x}{2 \times 3} sin (3x)}$

${= \frac {1}{9} \left [1 - \frac {D^2}{9} + \frac {D^4}{81} - \cdots \right ] x^2 + \frac {1}{9} \left [1 - \frac {D^2}{9} + \cdots \right ] x + \frac {x}{6} sin (3x)}$

All derivatives of ${x^2}$ of order ${\ge 3}$ will be ${0}$ and all derivatives of ${x}$ of order ${\ge 2}$ will be ${0}$. So,

${y_p = \frac {1}{9} \left [x^2 - \frac {2}{9} \right ] + \frac {1}{9} \left [2x \right ] + \frac {x \ sin (3x)}{6}}$

The complete solution is

${y = y_c + y_p = c_1 cos (3x) + c_2 sin (3x) + \frac {1}{9} \left [x^2 - \frac {2}{9} \right ] + \frac {1}{9} \left [2x \right ] + \frac {x \ sin (3x)}{6}}$