Solution of the differential equation $\frac{dy}{dx}+xy=xy^{2}$ is:

Question

Accepted Answer

To solve the differential equation $\frac{dy}{dx} + xy = xy^2$, we first rewrite it in a standard form suitable for solving:

$$
\frac{dy}{dx} = xy^2 - xy
$$

This can be rewritten as:

$$
\frac{dy}{dx} = xy(y - 1)
$$

This is a separable differential equation, which can be solved by separating the variables $y$ and $x$:

$$
\frac{dy}{y(y-1)} = x \, dx
$$

We can perform partial fraction decomposition on the left side:

$$
\frac{1}{y(y-1)} = \frac{A}{y} + \frac{B}{y-1}
$$

Solving for $A$ and $B$, we find:

$$
1 = A(y-1) + By
$$

Setting $y = 0$, we find $A = -1$, and setting $y = 1$, we find $B = 1$. Thus:

$$
\frac{1}{y(y-1)} = \frac{-1}{y} + \frac{1}{y-1}
$$

Substituting back, we have:

$$
\left(\frac{-1}{y} + \frac{1}{y-1}\right) dy = x \, dx
$$

Integrating both sides:

$$
\int \left(\frac{-1}{y} + \frac{1}{y-1}\right) dy = \int x \, dx
$$

This gives:

$$
-\ln|y| + \ln|y-1| = \frac{x^2}{2} + C
$$

Simplifying the left side:

$$
\ln\left|\frac{y-1}{y}\right| = \frac{x^2}{2} + C
$$

Exponentiating both sides:

$$
\left|\frac{y-1}{y}\right| = e^{\frac{x^2}{2} + C}
$$

Let $e^C = c$, a constant, then:

$$
\frac{y-1}{y} = \pm ce^{\frac{x^2}{2}}
$$

Assuming the positive case for simplicity:

$$
\frac{y-1}{y} = ce^{\frac{x^2}{2}}
$$

Rearranging gives:

$$
1 - \frac{1}{y} = ce^{\frac{x^2}{2}}
$$

Thus:

$$
-\frac{1}{y} = 1 + ce^{\frac{x^2}{2}}
$$

This matches Option 2: $-\frac{1}{y} = 1 + ce^{\frac{x^2}{2}}$.

**Explanation for Incorrect Options:**

1. **Option 1: $\frac{1}{y} = 1 + ce^{\frac{x^2}{2}}$**

This option is incorrect because the derived solution shows that the term on the left should be negative, as shown in the correct solution.

2. **Option 3: $\frac{1}{y} = 1 - ce^{x^2}$**

This option is incorrect because the exponent in the exponential function should be $\frac{x^2}{2}$, not $x^2$, and the sign of the constant term is incorrect.

3. **Option 4: None of these**

This option is incorrect because we have shown that Option 2 is indeed the correct solution to the differential equation.