Sample 2024-25 Class 12 Maths Q14 Solution: General solution of e^(y′) = x

The question

What is the general solution of the differential equation $e^{y'} = x$ ⁠?

(A) $y = x\log x + c$

(B) $y = x\log x - x + c$

(C) $y = x\log x + x + c$

(D) $y = x + c$

Exam answer

Taking log on both sides ( $x > 0$ ⁠),

e^{y'} = x \Rightarrow \frac{dy}{dx} = \log x

Why?

$\log$ (to base $e$ ⁠) undoes $e$ to a power: $\log e^u = u$ ⁠. With $u = y'$ this gives $y' = \log x$ ⁠, and $y'$ means $\frac{dy}{dx}$ ⁠.

Separating the variables and integrating,

\begin{aligned} &dy = \log x\,dx \\ \Rightarrow\ &\int dy = \int\log x\,dx \end{aligned}

y = \int\log x \cdot 1\,dx

Integrating by parts, with $\log x$ as the first function and 1 as the second function,

Why?

We cannot integrate $\log x$ directly, but we can differentiate it. So write it as $\log x \cdot 1$ and use integration by parts:

\begin{aligned} \int u\,v\,dx &= u\int v\,dx \\ &\quad - \int\left(\frac{du}{dx}\int v\,dx\right)dx \end{aligned}

Here $u = \log x$ ⁠, so $\frac{du}{dx} = \frac{1}{x}$ ⁠, and $v = 1$ ⁠, so $\int v\,dx = x$ ⁠.

\begin{aligned} y &= x\log x - \int x \cdot \frac{1}{x}\,dx \\ &= x\log x - \int 1\,dx \\ &= x\log x - x + c \end{aligned}

Why?

$x \cdot \frac{1}{x} = 1$ and $\int 1\,dx = x$ ⁠. One constant $c$ at the end is enough.

Check by differentiating:

For $x\log x$ use the product rule with $u = x$ ⁠, $v = \log x$ ⁠, $u' = 1$ ⁠, $v' = \frac1x$ ⁠:

\begin{aligned} \frac{d}{dx}(x\log x) &= u'v + uv' \\ &= (1)\log x + x \cdot \frac1x \\ &= \log x + 1 \end{aligned}

\begin{aligned} &\frac{d}{dx}(x\log x - x) \\ &= \log x + 1 - 1 = \log x \end{aligned}

Watch out: option (A) drops the $-x$ that comes from $\int 1\,dx$ ⁠.

Answer: (B) $y = x\log x - x + c$

Where this comes from

Chapter 9, Differential Equations: Differential equations: variables separable.

Question type: Variables separable + Integration by parts (combines 2 types)

How to spot it: click to see

Ch 9, type 3. Variables separable: $y$ on one side, $x$ on the other

Spot it: $\frac{dy}{dx}$ equals a part in $x$ times a part in $y$ ⁠, like $(1 + x^2)(1 + y^2)$ ⁠.

Here: Take log of both sides: $\frac{dy}{dx} = \log x$ ⁠, with no $y$ on the right. So $dy = \log x\,dx$ ⁠, and integrate both sides.

Ch 7, type 8. Integration by parts: a product of two different kinds

Spot it: a product such as $x\sin x$ ⁠, $xe^x$ or $x\log x$ ⁠. A lone $\log x$ or $\tan^{-1}x$ works too: take the other function as 1.

Here: $\int\log x\,dx$ has no card: write it as $\int\log x\cdot 1\,dx$ and integrate by parts, with $\log x$ first, to get $x\log x - x$ ⁠.

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Q14: General solution of ey′=xe^{y'} = xey′=x