Separate variables and solve dy/y = dx/x. General solution y = Cx.

Variable Separable — dy/dx = y/x

Question

Solve the differential equation:

\frac{dy}{dx} = \frac{y}{x}

Find the general solution.

Solution — Step by Step

The right-hand side has $y$ and $x$ in a clean ratio — that’s the signal to use variable separable. We’ll collect all $y$ terms on the left and all $x$ terms on the right.

Divide both sides so that $y$ terms are on the left and $x$ terms on the right:

\frac{dy}{y} = \frac{dx}{x}

This is valid as long as $y \neq 0$ and $x \neq 0$ — we’re essentially dividing both sides by $xy$ .

Now integrate both sides independently:

\int \frac{dy}{y} = \int \frac{dx}{x}

\ln|y| = \ln|x| + C_1

where $C_1$ is an arbitrary constant of integration.

Rewrite $C_1 = \ln|C|$ for some constant $C > 0$ (this is a standard trick — replacing an additive log constant with the log of a new constant):

\ln|y| = \ln|x| + \ln|C|

\ln|y| = \ln|Cx|

Removing the log from both sides:

y = Cx

General Solution: $y = Cx$ , where $C$ is an arbitrary constant.

Why This Works

The method works because we literally separated the equation into two independent integrals. Once $dy/y$ is on one side and $dx/x$ is on the other, neither side depends on the other variable — so we can integrate freely.

The step of writing $C_1 = \ln|C|$ is not magic — it just keeps the final answer cleaner. Instead of writing $e^{C_1} \cdot x$ , we absorb that exponential into a single constant $C$ . This is a technique worth practising because it appears in almost every variable separable problem in CBSE and JEE.

Geometrically, $y = Cx$ represents a family of straight lines through the origin. Each value of $C$ gives a different line — which makes sense, since differential equations have infinitely many solutions unless an initial condition pins one down.

Alternative Method — Using Homogeneous Substitution

This equation is also homogeneous (degree of numerator = degree of denominator = 1). We can substitute $y = vx$ , giving $dy/dx = v + x\,dv/dx$ .

Substituting:

v + x\frac{dv}{dx} = \frac{vx}{x} = v

x\frac{dv}{dx} = 0

\frac{dv}{dx} = 0 \implies v = C

Since $v = y/x$ , we get $y/x = C$ , so $y = Cx$ .

Both methods give the same answer here. For this particular equation, variable separable is faster. But if you see a ratio like $dy/dx = (x + y)/(x - y)$ , the homogeneous substitution becomes necessary.

Common Mistake

Forgetting the constant of integration on both sides.

Many students write:

\ln|y| = \ln|x| + C

…which is correct. But then they exponentiate carelessly to get $y = x + C$ . That’s wrong.

When you exponentiate $\ln|y| = \ln|x| + C$ , you get $y = e^C \cdot x$ , not $y = x + C$ . The $+C$ inside a logarithm becomes a multiplicative constant after removing the log — not an additive one.

The safe habit: always rewrite $e^C$ as a new constant $A$ or $C$ , and state $y = Cx$ clearly.

This exact question appeared in the CBSE 2024 Board Exam and is a standard 3-marker. The examiner expects you to show the separation step, the integration, and the log simplification explicitly — don’t skip steps in the interest of speed.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Homogeneous Substitution

Common Mistake

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