Q.The general solution of the differential equation $\frac{dy}{dx}=e^{\frac{x^2}{2}}+xy$ is:
(A) $y=ce^{-\frac{x^2}{2}}$
(B) $y=ce^{\frac{x^2}{2}}$
(C) $y=(x+c)e^{\frac{x^2}{2}}$
(D) $y=(c-x)e^{\frac{x^2}{2}}$

Concept: Integrating Factor Method — the equation is linear in $y$:

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

Step 1: Identify $P(x) = -x$, $Q(x) = e^{x^2/2}$.

Step 2: Integrating factor $\mu = e^{\int P\,dx} = e^{-\int x\,dx} = e^{-x^2/2}$.

Step 3: Multiply through:

$e^{-x^2/2}\frac{dy}{dx} - x e^{-x^2/2} y = 1$

$\Rightarrow \frac{d}{dx}\left(y e^{-x^2/2}\right) = 1$. …

The equation is $\frac{dy}{dx} = e^{x^2/2} + xy$. At first glance, it looks like it might be separable — but the $e^{x^2/2}$ term is added to $xy$, not multiplied by a function of $y$, so separation won't work directly. Instead, notice it's linear in $y$: we can rearrange it into the standard form $\frac{dy}{dx} - xy = e^{x^2/2}$.

Why does the Integrating Factor method work here? Because if we multiply the whole equation by a cleverly chosen function $\mu(x)$, the left-hand side becomes the derivative of $\mu(x) y$ — a perfect product rule. That turns the problem into a direct integration.

1. Rewrite in standard linear form Bring the $xy$ term to the left:

Here $P(x) = -x$ and $Q(x) = e^{x^2/2}$.

2. Find the Integrating Factor The formula is $\mu(x) = e^{\int P(x)\,dx}$.

So

3. Multiply through

The left side is exactly $\frac{d}{dx}\left( e^{-x^2/2} y \right)$ — check by differentiating: derivative of $e^{-x^2/2} y$ is $e^{-x^2/2} y' + y \cdot (-x e^{-x^2/2})$, which matches.

4. Integrate both sides

Integrate with respect to $x$:

5. Solve for $y$ Multiply through by $e^{x^2/2}$: …

Method: Integrating factor for a linear equation hidden by an added term

Use this when $\frac{dy}{dx}$ equals (a function of $x$) plus a term like $xy$ — linear in $y$, but not separable because the extra piece is added, not multiplied.

Steps

Step 1: Rearrange to $\frac{dy}{dx}+P(x)y=Q(x)$.

Bring the $xy$ term to the left; here $P(x)=-x$ and $Q(x)$ is the leftover $x$-function.

Step 2: Build the integrating factor.

$\text{I.F.}=e^{\int P\,dx}.$ …

Common Mistakes

Mistake 1: Trying to separate the variables.

Why it's wrong: in $\frac{dy}{dx}=e^{x^2/2}+xy$ the $xy$ term is added, not a factor, so the equation does not separate. Correct approach: rearrange to the linear form $\frac{dy}{dx}-xy=e^{x^2/2}$ and use the I.F.

Mistake 2: Sign error in $P(x)$. …