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}}

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