Board 2025 Class 12 Maths Q12 Solution: Integrating factor of (x + 2y³) dy/dx = 2y

The question

The integrating factor of differential equation $(x + 2y^3)\frac{dy}{dx} = 2y$ is

(A) $e^{\frac{y^2}{2}}$

(B) $\frac{1}{\sqrt{y}}$

(C) $\frac{1}{y^2}$

(D) $e^{-\frac{1}{y^2}}$

Exam answer

$\frac{dy}{dx} = \frac{2y}{x + 2y^3}$ is not linear in y. Taking the reciprocal,

Why?

A linear equation in y has the form $\frac{dy}{dx} + Py = Q$ ⁠, with P and Q functions of x only. Here $y^3$ sits in the denominator, so that form is impossible.

But in $\frac{dx}{dy}$ ⁠, x appears only to the first power. So treat y as the variable and x as the unknown function.

\begin{aligned} \frac{dx}{dy} &= \frac{x + 2y^3}{2y} \\ &= \frac{x}{2y} + y^2 \\ \Rightarrow \frac{dx}{dy} - \frac{1}{2y}\,x &= y^2 \end{aligned}

This is linear in x, of the form $\frac{dx}{dy} + Px = Q$ ⁠, with $P = -\frac{1}{2y}$ and $Q = y^2$ ⁠.

Why?

For $\frac{dx}{dy} + Px = Q$ ⁠, with P, Q functions of y:

\text{I.F.} = e^{\int P\,dy}

Taking the reciprocal turns $\frac{2y}{x + 2y^3}$ upside down. Then split the fraction into two:

\begin{aligned} \frac{x + 2y^3}{2y} &= \frac{x}{2y} + \frac{2y^3}{2y} \\ \frac{2y^3}{2y} &= y^{3 - 1} = y^2 \end{aligned}

Subtract $\frac{x}{2y}$ from both sides, and write $\frac{x}{2y} = \frac{1}{2y}\,x$ ⁠.

Watch out for the sign: moving $\frac{x}{2y}$ to the left makes P negative.

\begin{aligned} \text{I.F.} &= e^{\int -\frac{1}{2y}\,dy} \\ &= e^{-\frac{1}{2}\log y} = \frac{1}{\sqrt{y}} \end{aligned}

Why?

$\int \frac{1}{y}\,dy = \log y$ (here $y > 0$ ⁠), so $\int -\frac{1}{2y}\,dy = -\frac12\log y$ ⁠.

Take the constant out: $\int -\frac{1}{2y}\,dy = -\frac12\int\frac{1}{y}\,dy$ ⁠. Move the $-\frac12$ inside the log as a power, using $k\log y = \log y^k$ ⁠. Then use $e^{\log u} = u$ ⁠:

\begin{aligned} e^{-\frac12\log y} &= e^{\log y^{-\frac12}} \\ &= y^{-\frac12} = \frac{1}{\sqrt{y}} \end{aligned}

Answer: (B) $\dfrac{1}{\sqrt{y}}$

Where this comes from

Chapter 9, Differential Equations: Integrating factor of a linear DE in x.

Question type: Linear in $x$

How to spot it: click to see

Type 8. Linear in $x$ ⁠: when $y$ will not behave, solve for $x$

Spot it: there is a $y^2$ or $\tan^{-1}y$ next to $dy$ ⁠, but $x$ appears only to power 1.

Here: The $2y^3$ beside $x$ stops it being linear in $y$ ⁠, but $x$ is only to power 1. Flip to $\frac{dx}{dy} - \frac{1}{2y}x = y^2$ ⁠, so $\text{I.F.} = e^{-\frac12\log y} = \frac{1}{\sqrt y}$ ⁠.

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Q12: Integrating factor of (x+2y3)dydx=2y(x + 2y^3)\frac{dy}{dx} = 2y(x+2y3)dxdy​=2y

Q12: Integrating factor of $(x + 2y^3)\frac{dy}{dx} = 2y$