A curve passes through the point (x = 1, y = 0) and satisfies the
A. <span class="math-tex">\({\rm{In}}\left( {1 + \frac{{{{\rm{y}}^2}}}{{{{\rm{x}}^2}}}} \right) = {\rm{x}} - 1\)</span>
B. <span class="math-tex">\(\frac{1}{2}{\rm{In}}\left( {1 + \frac{{{{\rm{y}}^2}}}{{{{\rm{x}}^2}}}} \right) = {\rm{x}} - 1\)</span>
C. <span class="math-tex">\({\rm{In}}\left( {1 + \frac{{\rm{y}}}{{\rm{x}}}} \right) = {\rm{x}} - 1\)</span>
D. <span class="math-tex">\(\frac{1}{2}{\rm{In}}\left( {1 + \frac{{\rm{y}}}{{\rm{x}}}} \right) = {\rm{x}} - 1\)</span>
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Right Answer is: A
SOLUTION
\(\frac{{dy}}{{dx}} = \frac{{{x^2} + {y^2}}}{{2y}} + \frac{y}{x}\)
\(\frac{{dy}}{{dx}} = \frac{{{x^2}}}{{2y}} + \frac{y}{2} + \frac{y}{x}\)
\(y\frac{{dy}}{{dx}} - \left( {\frac{1}{x} + \frac{1}{2}} \right){y^2} = \frac{{{x^2}}}{2}\)
y2 = t
\(2y\frac{{dy}}{x} = \frac{{dt}}{{dx}}\)
\(\frac{1}{2}\frac{{dt}}{{dx}} - \left( {\frac{1}{x} + \frac{1}{2}} \right)t = \frac{{{x^2}}}{2}\)
\(\frac{{dt}}{{dx}} - 2\left( {\frac{1}{x} + \frac{1}{2}} \right)t = {x^2}\)
The above is linear differential equation of the form
\(\frac{{dt}}{{dx}} = + pt = Q\)
Integrating factor \(= {e^{ - 2\smallint \left( {\frac{1}{x} + \frac{1}{2}} \right)dx}}\)
= e-2ln x. e-x
\(= {e^{\ln {x^{ - 2}}}}.{e^{ - x}}\)
\(= \frac{{{e^{ - x}}}}{{{x^2}}}\)
Solution is given by:
\(\frac{{t\;{e^{ - x}}}}{{{x^2}}} = \smallint \frac{{{x^2}{e^{ - x}}}}{{{x^2}}}dx + c\)
Substitute t = y2
\(\frac{{{y^2}{e^{ - x}}}}{{{x^2}}} = - {e^{ - x}} + c\)
At, x = 1, y = 0
0 = -e-1 + c
c = e-1 = ye
substituting in the solution
\({y^2}.\frac{{{e^{ - x}}}}{{{x^2}}} = - {e^{ - x}} + {e^{ - 1}}\)
\(\left( {\frac{{{y^2}}}{{{x^2}}} + 1} \right){e^{ - x}} = {e^{ - 1}}\)
\(\left( {1 + \frac{{{y^2}}}{{{x^2}}}} \right) = \frac{{{e^{ - 1}}}}{{{e^{ - x}}}} = {e^{x - 1}}\)
Taking log both sides
\(\ln \left( {1 + \frac{{{y^2}}}{{{x^2}}}} \right) = x - 1\)