1 markMCQSection A

General solution of differential equation $y \log y \, dx - x \, dy = 0$ is :

Q: General solution of differential equation $y \log y \, dx - x \, dy = 0$ is :

The correct option is **B) $y = e^{|cx|}$** (i.e. $y = e^{cx}$). **Working:** - $y \log y \, dx - x \, dy = 0 \Rightarrow y \log y \, dx = x \, dy$. - Separate variables: $\frac{dx}{x} = \frac{dy}{y \log y}$. - Integrate: $\int \frac{dx}{x} = \int \frac{dy}{y \log y}$. Put $u = \log y$, $du = \frac{dy}{y}$, so the RHS = $\int \frac{du}{u}$. - $\log|x| = \log|\log y|$ + constant ⇒ $\log y = c \cdot x$. - Therefore $y = e^{cx}$.

Q: Why is the constant multiplicative ($cx$) and not additive ($c + x$)?

Integration gives $\log|x| = \log|\log y| + c_1$. Exponentiating both sides moves the additive constant into a multiplicative factor, so $\log y = cx$, not $c + x$.

Q: What substitution simplifies $\int \frac{dy}{y \log y}$?

Let $u = \log y$, then $du = \frac{dy}{y}$ and the integral becomes $\int \frac{du}{u} = \log|u| = \log|\log y|$.

Calculus (Differential Equations)

Differential Equations — variable separable

Options

(A)

y = \log |cx|

(B)

y = e^{|cx|}

(C)

y = e^{c + x}

(D)

\log y = |c + x|

Official Answer

The correct option is B) $y = e^{|cx|}$ (i.e. $y = e^{cx}$ ).

Working:

$y \log y \, dx - x \, dy = 0 \Rightarrow y \log y \, dx = x \, dy$ .
Separate variables: $\frac{dx}{x} = \frac{dy}{y \log y}$ .
Integrate: $\int \frac{dx}{x} = \int \frac{dy}{y \log y}$ . Put $u = \log y$ , $du = \frac{dy}{y}$ , so the RHS = $\int \frac{du}{u}$ .
$\log|x| = \log|\log y|$ + constant ⇒ $\log y = c \cdot x$ .
Therefore $y = e^{cx}$ .

variable separabledifferential equationy log ysubstitution u = log ylog y = cxy = e^{cx}general solution

Marking Scheme

11 mark: correct separation, integration to $\log y = cx$ , and answer $y = e^{cx}$ (option B).

Hint

Separate to $\frac{dx}{x} = \frac{dy}{y \log y}$ ; substitute $u = \log y$ so the right side is $\int \frac{du}{u}$ .

Quick Oral Answer

Separating gives $\frac{dx}{x} = \frac{dy}{y \log y}$ ; with $u = \log y$ this integrates to $\log y = cx$ , so $y = e^{cx}$ .

Analysis & Explanation

This is a classic variable-separable differential equation with a substitution on the y-side.

Concept:

Separate x-terms and y-terms, then integrate each side.
The substitution $u = \log y$ turns $\int \frac{dy}{y \log y}$ into the standard $\int \frac{du}{u} = \log|u|$ .

Why B is correct:

Integration gives $\log|x| = \log|\log y| + c_1$ , i.e. $\log y = cx$ , hence $y = e^{cx}$ , matching option B.

Why the distractors are wrong:

A ( $y = \log|cx|$ ): inverts the exponential relation incorrectly.
C ( $y = e^{c+x}$ ): comes from treating the constant additively ( $\log y = c + x$ ) instead of multiplicatively.
D ( $\log y = |c + x|$ ): again uses an additive constant, which does not follow from $\log|x| = \log|\log y| + c$ .

Common Mistakes

1Taking the arbitrary constant additively ( $\log y = c + x$ ) instead of multiplicatively ( $\log y = cx$ ).
2Forgetting the substitution $u = \log y$ and mis-integrating $\int \frac{dy}{y \log y}$ .
3Dropping the exponential step, leaving the answer as $\log y$ instead of y.

Interesting Facts

The integral $\int \frac{dx}{x \log x} = \log|\log x|$ is a textbook example of a nested-log substitution.

Equations of the form $\frac{dy}{dx} = y \cdot f(x)$ always yield exponential-type solutions, $y = e^{\int f\,dx}$ .

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Frequently Asked Questions

Why is the constant multiplicative ( $cx$ ) and not additive ( $c + x$ )?

Integration gives $\log|x| = \log|\log y| + c_1$ . Exponentiating both sides moves the additive constant into a multiplicative factor, so $\log y = cx$ , not $c + x$ .

What substitution simplifies $\int \frac{dy}{y \log y}$ ?

Let $u = \log y$ , then $du = \frac{dy}{y}$ and the integral becomes $\int \frac{du}{u} = \log|u| = \log|\log y|$ .