If y' = x - y x + y , then its solution is - Vidyadip

MCQ

If $y' = \frac{{x - y}}{{x + y}}$, then its solution is

✓
${y^2} + 2xy - {x^2} = c$
B
${y^2} + 2xy + {x^2} = c$
C
${y^2} - 2xy - {x^2} = c$
D
${y^2} - 2xy + {x^2} = c$

✓

Answer

Correct option: A.

${y^2} + 2xy - {x^2} = c$

a
(a) Given $\frac{{dy}}{{dx}} = \frac{{x - y}}{{x + y}}$. Put $y = vx$ ==> $\frac{{dy}}{{dx}} = v + x\,\frac{{dv}}{{dx}}$

$v + x\,\frac{{dv}}{{dx}} = \frac{{x - vx}}{{x + vx}}$

==> $v + x\,\frac{{dv}}{{dx}} = \frac{{1 - v}}{{1 + v}}$ ==> $\frac{{1 + v}}{{2 - {{(1 + v)}^2}}}dv = \frac{{dx}}{x}$

Integrating both sides, $\int {\frac{{1 + v}}{{2 - {{(1 + v)}^2}}}} \,dv = \int {\frac{{dx}}{x}} $

Put ${(1 + v)^2} = t \Rightarrow 2(1 + v)dv = dt$

==> $\frac{1}{2}\int_{}^{} {\frac{{dt}}{{2 - t}}} = \int_{}^{} {\frac{{dx}}{x}} $ ==> $ - \frac{1}{2}\log (2 - t) = \log xc$

==> $ - \frac{1}{2}\log [2 - {(1 + v)^2}] = \log xc$

==> $ - \frac{1}{2}\log [ - {v^2} - 2v + 1] = \log xc$

==> $\log \frac{1}{{\sqrt {1 - 2v - {v^2}} }} = \log xc$

==> ${x^2}{c^2}(1 - 2v - {v^2}) = 1$ ==> ${y^2} + 2xy - {x^2} = {c_1}$.

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