Solve: $x\frac{dy}{dx}=y(\log y-\log x+1)$.

Solve: x(dy)/(dx)=y(log y-log x+1).

NCERT Exemplar · Q33

x\frac{dy}{dx}=y(\log y-\log x+1)

Yanam CbseLong· 5mImportance★★★★★

Appeared in past exams:MHT-CET 2025· Set pcm-2025-04-23-M· 2mexactMHT-CET 2024· Set pcm-2024-05-02-M· 2mexact

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Concept understanding — Separation Of Variables

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:

\frac{dy}{dx} = f(x) \cdot g(y)

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:

\frac{dy}{dx} = f(x) \cdot g(y) \quad \Longrightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx

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:

\int \frac{1}{g(y)} \, dy = \int f(x) \, dx

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

Watch out

You cannot separate if the equation is not in product form. For example, $\frac{dy}{dx} = x + y$ cannot be separated — the sum $x+y$ is not a product $f(x)g(y)$ .

Why This Works

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

\frac{1}{g(y)} \frac{dy}{dx} = f(x)

Now integrate both sides with respect to $x$ :

\int \frac{1}{g(y)} \frac{dy}{dx} \, dx = \int f(x) \, dx

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 $y \neq 0$ ):

\frac{1}{y} \, dy = 2x \, dx

Step 2: Integrate.

\int \frac{1}{y} \, dy = \int 2x \, dx \quad\Longrightarrow\quad \log|y| = x^2 + C

Step 3: Solve for $y$ .

|y| = e^{x^2 + C} = e^C \cdot e^{x^2}

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: $\boxed{y = 3e^{x^2}}$ …

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This is a homogeneous differential equation that becomes separable after substituting $y = vx$ . The solution is $\log\left(\frac{y}{x}\right) = Cx$ .

The key insight here is recognising the structure. When you see terms like $\log y - \log x$ , your first thought should be to combine them: $\log(y/x)$ . That immediately suggests the substitution $y = vx$ , because $y/x = v$ turns the logarithm into something simple. This is the classic move for homogeneous differential equations — equations where every term has the same total degree in $x$ and $y$ .

Let’s check homogeneity. The left side $x \frac{dy}{dx}$ is degree 1 in $x$ and $y$ (since $dy/dx$ is degree 0). The right side $y(\log(y/x) + 1)$ : $y$ is degree 1, and the bracket depends only on the ratio $y/x$ , so it’s degree 0. The whole right side is degree 1. Homogeneous — good.

Now the substitution $y = vx$ will reduce it to a separable equation in $v$ and $x$ .

Substitute $y = vx$

Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$ by the product rule. Also $\log y - \log x = \log(vx) - \log x = \log v$ .

The equation becomes:

x\left(v + x\frac{dv}{dx}\right) = vx(\log v + 1)

Simplify Cancel $x$ (assuming $x \neq 0$ ):

v + x\frac{dv}{dx} = v(\log v + 1)

Expand the right: $v\log v + v$ .

Isolate the derivative Subtract $v$ from both sides:

x\frac{dv}{dx} = v\log v

This is now separable — all $v$ terms on one side, all $x$ on the other.

Separate variables

\frac{dv}{v\log v} = \frac{dx}{x}

Watch out

A common mistake here is forgetting the absolute value inside the log when integrating. Always write $\log|v|$ and $\log|x|$ until you know the sign. Also, $v=0$ or $v=1$ are special cases — $v=0$ gives $y=0$ , which trivially satisfies the original equation? Check: if $y=0$ , the right side becomes $0(\log 0 - \log x + 1)$ which is undefined. So $v=0$ is not allowed. $v=1$ gives $\log v = 0$ , making the left side zero — that’s a valid constant solution.

Integrate both sides …

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Q.Solve: $x\frac{dy}{dx}=y(\log y-\log x+1)$ .

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

Method: Homogeneous DE with a logarithmic ratio ( $v=y/x$ )

Steps

Common Mistakes

Q.Solve: xdydx=y(log⁡y−log⁡x+1)x\frac{dy}{dx}=y(\log y-\log x+1)xdxdy​=y(logy−logx+1).

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Q.Solve: $x\frac{dy}{dx}=y(\log y-\log x+1)$ .

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