Q.The solution of the equation $(2y-1)\,dx-(2x+3)\,dy=0$ is:
(A) $\frac{2x-1}{2y+3}=k$
(B) $\frac{2y+1}{2x-3}=k$
(C) $\frac{2x+3}{2y-1}=k$
(D) $\frac{2x-1}{2y-1}=k$

Separation of Variables: From Intuition to Precision

Imagine you're baking a cake. The recipe says "mix the dry ingredients separately, then add the wet ones." You keep things that belong together together, and things that don't apart — until the right moment. Separation of Variables does exactly that for certain kinds of equations.

The Core Intuition

Some equations involve two different kinds of change happening at once. Think of a cup of hot coffee cooling down. The rate at which it cools depends on:

• The temperature difference between the coffee and the room (a function of time)
• The surface area of the cup (a function of shape, not time)

These two influences are tangled together in one equation. Separation of Variables is the mathematical trick that untangles them — it lets you handle the time part first, then the space part separately.

The Precise Statement

Separation of Variables applies to ordinary differential equations (ODEs) of the form:

where the right-hand side is a product of a function of $x$ alone and a function of $y$ alone. The method works in three clean steps:

Step 1: Separate. Multiply both sides by $dx$ and divide by $g(y)$ (assuming $g(y) \neq 0$). This moves all $y$'s to one side and all $x$'s to the other.

Step 2: Integrate. Put an integral sign on both sides:

Step 3: Solve. Evaluate both integrals and solve for $y$ explicitly if possible.

Why This Works

The justification is the chain rule in reverse. From $\frac{dy}{dx} = f(x)g(y)$, rewrite it as:

Now integrate both sides with respect to $x$:

The left side is a substitution waiting to happen: $\frac{dy}{dx} dx = dy$, so you get $\int \frac{1}{g(y)} dy$. That's the entire trick — the chain rule dressed up.

A Concrete Example

Solve $\frac{dy}{dx} = 2xy$, with $y(0) = 3$.

Step 1: Separate. Divide both sides by $y$ (assuming $y \neq 0$):

Step 2: Integrate.

Step 3: Solve for $y$.

Let $A = \pm e^C$ (absorbing the absolute value): $y = A e^{x^2}$. Now use $y(0) = 3$: $3 = A e^{0} = A$, so $A = 3$.

Final answer: $\boxed{y = 3e^{x^2}}$ …