Right choice is (c) dθ = \(\sqrt {M^2 – 1} \frac {DV}{V}\)

Explanation: If we set the numerator of the combination of momentum, continuity and energy equation REPRESENTED by Cramer’s RULE as zero, we get

(1 – \(\frac {u^{2}}{a^{2}}\))dudy + (1 – \(\frac {v^{2}}{a^{2}}\))dxdv = 0

On rearranging the terms we get

(1 – \(\frac {v^{2}}{a^{2}}\))dxdv = – (1 – \(\frac {u^{2}}{a^{2}}\))dudy

\(\frac {dv}{du} = \frac { – (1 – \frac {u^{2}}{a^{2}})dy}{(1 – \frac {v^{2}}{a^{2}})DX}\)

Substituting the value of characteristic curve in the above equation –

\(\frac {dy}{dx_{char}} =\frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}} ) – 1}}{1 – \frac {u^{2}}{a^{2}}}\)

We get, \(\frac {dv}{du} = \frac{- (1 – \frac {u^{2}}{a^{2}})}{(1 – \frac {v^{2}}{a^{2} })} \big [ \frac { – \frac {uv}{a^{2}} ± \sqrt {(\frac {u^{2} + v^{2}}{a^{2}} ) – 1}}{1 – \frac {u^{2}}{a^{2}}} \big ] = \frac {\frac {uv}{a^{2}} ∓ \sqrt {(\frac {u^{2} + v^{2}}{a^{2}} ) – 1}}{1 – \frac {v^{2}}{a^{2}}}\)

u and v are the x and y – component of velocity V. u = Vcosθ and v = Vsinθ. On substituting these VALUES we get

\(\frac {d(Vsinθ)}{d(Vcosθ)} = \frac {M^2 cosθsinθ ∓ \sqrt {M^2 – 1} }{1 – M^2 sin^2 θ}\)

dθ = ∓ \(\sqrt {M^2 – 1}\frac {dV}{V}\)

This the characteristic line along the C+ characteristic line is .dθ = + \(\sqrt {M^2 – 1}\frac {dV}{V}\).