Q.Find the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ given that $y = 1$, when $x = 0$.

Concept: Separation of Variables

We rewrite the equation as

$\frac{dy}{y^2} = -4x \, dx$.

Integrate both sides:

$\int y^{-2} dy = \int -4x \, dx$

$\Rightarrow -y^{-1} = -2x^2 + C$.

Use the initial condition $y=1$ when $x=0$:

$-1 = 0 + C \Rightarrow C = -1$. …

The equation $\frac{dy}{dx} = -4xy^2$ is a classic example of separation of variables. The idea is simple: if you can rewrite the derivative so that all the $y$'s are on one side of the equation and all the $x$'s are on the other, you can integrate each side separately. Here, the right-hand side is a product of a function of $x$ ($-4x$) and a function of $y$ ($y^2$), which is the perfect setup.

Why does this work? Because we treat $dy/dx$ as a fraction (carefully, in the context of differentials) and rearrange. Then integrating both sides with respect to their own variable recovers the relationship between $y$ and $x$. The constant of integration is then pinned down by the given condition $y=1$ when $x=0$.

Let’s walk through it.

1. Separate the variables. Multiply both sides by $dx$ and divide by $y^2$ (assuming $y \neq 0$, which is fine since $y=1$ at the start):

2. Integrate both sides. The left side integrates with respect to $y$, the right with respect to $x$:

Simplify the right side:

3. Solve for $y$ in terms of $x$. …

Method: Separable equation with an initial condition (particular solution)

Use this when you are given a separable equation together with a data point such as "$y = y_0$ when $x = x_0$" and asked for the particular (not general) solution.

Steps

Step 1: Separate and integrate to get the general solution.

Move all $y$ to one side, all $x$ to the other, integrate, and keep the arbitrary constant:

Step 2: Substitute the given point to pin down $C$. …

Common Mistakes

Mistake 1: Wrong antiderivative of $y^{-2}$.

Why it's wrong: $\displaystyle\int y^{-2}\,dy = -y^{-1}$, not $y^{-1}$ or $\dfrac{y^{-3}}{3}$. A sign or power slip here derails the rest. Correct approach: check by differentiating — $\dfrac{d}{dy}(-y^{-1}) = y^{-2}$.

Mistake 2: Applying the initial condition too early.

Why it's wrong: substituting $y=1,\,x=0$ before integrating leaves nothing to determine and loses the constant. Correct approach: get the general solution first, then plug in the point to find $C$. …