If dy dx = 1 + x + y + xy and y( - 1) = 0, then function y is - Vidyadip

MCQ

If $\frac{{dy}}{{dx}} = 1 + x + y + xy$ and $y( - 1) = 0,$ then function $y$ is

A
${e^{{{(1 - x)}^2}/2}}$
✓
${e^{{{(1 + x)}^2}/2}} - 1$
C
${\log _e}(1 + x) - 1$
D
$1 + x$

✓

Answer

Correct option: B.

${e^{{{(1 + x)}^2}/2}} - 1$

b
(b) $\frac{{dy}}{{dx}} = 1 + x + y + xy$
==> $\frac{{dy}}{{dx}} = (1 + x) + y{\rm{ }}(1 + x)$
==> $\frac{{dy}}{{dx}} = (1 + x)(1 + y)$
==> $\frac{{dy}}{{(1 + y)}} = dx(1 + x)$
Integrating both sides,$\int_{}^{} {\frac{{dy}}{{(1 + y)}}} = \int_{}^{} {dx(1 + x)} $
$\log (1 + y) = x + \frac{{{x^2}}}{2} + \log c$
$y = c{e^{x + ({x^2}/2)}} - 1$
==> $y( - 1) = c{e^{ - 1 + (1/2)}} - 1 = 0$
$\therefore $$c{e^{ - 1/2}} = 1$ ==> $c = {e^{1/2}}$
$\therefore y = {e^{1/2}}{e^{x + \frac{{{x^2}}}{2}}} - 1$, $y = {e^{\frac{{{{(x + 1)}^2}}}{2}}} - 1$.

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