The general Solution of differential equation dx = e 2 + xy is

Given that, dx = e x 2 /2 + xy dx - xy = e x 2 /2 Here, P = - x,Q = e x 2 /2 IF = e - xdx = e - x 2 /2 The general Solution is y /2 = /2 - e x 2 /2 dx + C y e - x 2 /2 = 1 dx + C y /2 = x + C y = x e x 2 /2 + C e + x 2 /2 y = (x + C) e x 2 /2

Differential Equations — Class 12 Maths Solution

exemplar objective MCQ NCERT EXEMP.Q.63,Page.199

Question

The general Solution of differential equation $\frac{{dy}}{{dx}} = {e^{\frac{{{x^2}}}{2}}} + xy$ is

(a) $y = C{e^{ - {x^2}/2}}$
(b) $y = C{{\rm{e}}^{{x^2}/2}}$
(c) $y = (x + C){e^{{x^2}/2}}$ ✓ Correct
(d) $y = (C - x){e^{{x^2}/2}}$

Step-by-step Solution

Given that, $\frac{{dy}}{{dx}} = {e^{{x^2}/2}} + xy$

$\Rightarrow$ $\frac{{dy}}{{dx}} - xy = {e^{{x^2}/2}}$

Here, $P = - x,Q = {e^{{x^2}/2}}$

$\therefore$ $IF = {e^{\int - xdx}} = {e^{ - {x^2}/2}}$

The general Solution is
$y \cdot {e^{ - {x^2}/2}} = \int {{e^{ - {x^2}/2}}} - {e^{{x^2}/2}}dx + C$

$\Rightarrow$ $y{e^{ - {x^2}/2}} = \int 1 dx + C$

$\Rightarrow$ $y \cdot {e^{ - {x^2}/2}} = x + C$

$\Rightarrow$ $y = x{e^{{x^2}/2}} + C{e^{ + {x^2}/2}}$

$\Rightarrow$ $y = (x + C){e^{{x^2}/2}}$

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