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Akanksha P.

Government teacher

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Linear Algebra

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Question:

If $ W = \left\{ {y:y\,is\,\,a\,real\,valued\,function\,on\,{\rm{real no}}.{\rm{ :}}\frac{{{d^2}y}}{{d{x^2}}} - 5\frac{{dy}}{{dx}} + 6y = 0} \right\}$ What is the dimesion of $W$.

Akanksha P.

Answer:

$W$ forms vector space . Let $\frac{d}{{dx}} = D$ So \[\begin{array}{l} \left( {{D^2} - 5D + 6} \right)y = 0\\ \left( {D - 2} \right)\left( {D - 3} \right)y = 0\\ y = \alpha {e^{2x}} + \beta {e^{3x}} \end{array}\] where $\alpha and \beta $ are arbitrary constants. Here ${e^{2x}},{e^{3x}}$ are two linearly independent solution of D.E. Then $dim(W)=2$.

Algebra

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Question:

If \[\frac{{{Q_8}}}{{Z({Q_8})}} \cong G\] then $G$ is \[\begin{array}{l} i){Q_8}\\ ii){Z_2}\\ iii){K_4}\,\,or\,\,{Z_2} \times {Z_2}\\ iv){Z_4} \end{array}\] where \[{Q_8} = {\rm{Hamiltonian}}\,\,{\rm{group}}\]${Z_m}\,$ represents cyclic group of order $m$ and ${K_4}\,$ is Kilien's four group.

Akanksha P.

Answer:

As we know that \[{Q_8} = \left\{ { \pm 1, \pm i, \pm j, \pm k} \right\}\,\]and\[Z({Q_8}) = \left\{ { \pm 1} \right\}\,\] So order of ${Q_8} = 8\,\,{\rm{and order}}\,{\rm{of}}\,Z({Q_8}) = 2$ i.e.,order of $\frac{{{Q_8}}}{{Z({Q_8})}} = 4$ Now we have only two possibilities Let us consider $G = {Z_4}$,$Z_4}$ is a cyclic group of order 4. It means $\frac{{{Q_8}}}{{Z({Q_8})}}$ is cyclic. We have a theorem,\[If\,\frac{G}{{Z(G)}}\,is\,cyclic \Rightarrow G\,is\,abelian.\] which imply${{Q_8}}$ is abelian. This is wrong. Hence $G = {K_4}$.

Partial Differential Equations

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Question:

Find the complete integral of Partial Differential Equation \begin{equation} 6z\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\frac{{\partial z}}{{\partial y}}} \right) = 3{\left( {\frac{{\partial z}}{{\partial x}}} \right)^2}\left[ {3{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^3} + 2x\left( {\frac{{\partial z}}{{\partial y}}} \right)} \right] + 2{\left( {\frac{{\partial z}}{{\partial y}}} \right)^2}\left[ {2{{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^3} + 3y\left( {\frac{{\partial z}}{{\partial x}}} \right)} \right] \end{equation}

Akanksha P.

Answer:

For simplification, Let us consider \[p = \left( {\frac{{\partial z}}{{\partial x}}} \right),and q = \left( {\frac{{\partial z}}{{\partial y}}} \right)\] so we can write the given equation as \[6zpq = 3{p^2}\left[ {3{p^3} + 2xq} \right] + 2{q^2}\left[ {2{q^3} + 3yp} \right]\] i.e.,\[\begin{array}{l} z = \frac{{3{p^2}\left[ {3{p^3} + 2xq} \right] + 2{q^2}\left[ {2{q^3} + 3yp} \right]}}{{6pq}}\\ z = \frac{{3{p^2}\left[ {3{p^3} + 2xq} \right]}}{{6pq}} + \frac{{2{q^2}\left[ {2{q^3} + 3yp} \right]}}{{6pq}}\\ z = \frac{3}{2}\frac{{{p^4}}}{q} + px + \frac{2}{3}{q^4} + qy \end{array}\] Now it becomes Clairates Equation So the complete integral of Partial Differential Equation is \[z = \frac{3}{2}\frac{{{a^4}}}{q} + ax + \frac{2}{3}{b^4} + by\] where $a$ and $b$ are arbitrary constants.

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