CBSE Board Examination 2026 · Section C · 3 marks

Q29: Solve $x^2\frac{dy}{dx} = x^2 + xy + y^2$ ⁠, or $xy\frac{dy}{dx} = (x + 2)(y + 2)$ with $y(1) = -1$

Board 2026Ch 9 · Differential Equations

The question

(a) Find the general solution of the following differential equation :

x^2\frac{dy}{dx} = x^2 + xy + y^2

(b) Find the particular solution of the differential equation

xy\frac{dy}{dx} = (x + 2)(y + 2),

given that $y(1) = -1$ ⁠.

Exam answer

(a)

Dividing by $x^2$ ( $x \ne 0$ ⁠), the given equation is

\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2

Why?

Each term on the right has degree 2 in $x$ and $y$ together ( $x^2$ ⁠, $xy$ ⁠, $y^2$ ⁠), so the equation is homogeneous. Dividing by $x^2$ shows the right side as a function of $\frac{y}{x}$ alone:

\frac{xy}{x^2} = \frac{y}{x},\quad \frac{y^2}{x^2} = \left(\frac{y}{x}\right)^2

It is homogeneous. Put $y = vx$ ⁠. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$ ⁠.

Why?

For a homogeneous equation the substitution $y = vx$ always separates the variables.

$v$ depends on $x$ ⁠, so use the product rule on $vx$ ⁠:

\frac{d}{dx}(vx) = v\cdot 1 + x\frac{dv}{dx}

\begin{aligned} v + x\frac{dv}{dx} &= 1 + v + v^2 \\ \Rightarrow x\frac{dv}{dx} &= 1 + v^2 \\ \Rightarrow \frac{dv}{1 + v^2} &= \frac{dx}{x} \end{aligned}

Why?

Put $\frac{y}{x} = v$ on the right. The $v$ on both sides cancels.

\begin{aligned} v + x\frac{dv}{dx} &= 1 + v + v^2 \\ x\frac{dv}{dx} &= 1 + v + v^2 - v \\ x\frac{dv}{dx} &= 1 + v^2 \end{aligned}

Then separate: everything with $v$ goes with $dv$ ⁠, everything with $x$ goes with $dx$ ⁠.

Divide both sides by $1 + v^2$ and by $x$ ⁠, and multiply by $dx$ ⁠:

\frac{dv}{1 + v^2} = \frac{dx}{x}

Integrating both sides,

\begin{aligned} \int\frac{dv}{1 + v^2} &= \int\frac{dx}{x} \\ \tan^{-1}v &= \log|x| + C \end{aligned}

Why?

Standard integrals:

\begin{aligned} \int\frac{dv}{1 + v^2} &= \tan^{-1}v \\ \int\frac{dx}{x} &= \log|x| \end{aligned}

One arbitrary constant $C$ is enough for the two sides.

Putting $v = \frac{y}{x}$ ⁠,

Why?

The answer must be in the original variables, so replace $v$ by $\frac{y}{x}$ ⁠.

$\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + C$ is the general solution.

(b)

Separating the variables ( $x \ne 0$ ⁠, $y \ne -2$ ⁠),

xy\,dy = (x + 2)(y + 2)\,dx

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

Why?

The right side is a product of a function of $x$ and a function of $y$ ⁠, so the variables separate. Divide both sides by $x(y + 2)$ ⁠: the $y$ -terms stay with $dy$ ⁠, the $x$ -terms with $dx$ ⁠.

First, multiply both sides of the question by $dx$ ⁠:

xy\,dy = (x + 2)(y + 2)\,dx

Then divide both sides by $x(y + 2)$ ⁠:

\begin{aligned} \text{Left: } &\frac{xy}{x(y + 2)}\,dy = \frac{y}{y + 2}\,dy \\[4pt] \text{Right: } &\frac{(x + 2)(y + 2)}{x(y + 2)}\,dx \\ &= \frac{x + 2}{x}\,dx \end{aligned}

Integrating both sides,

\begin{aligned} &\int\left(1 - \frac{2}{y + 2}\right)dy \\ &= \int\left(1 + \frac{2}{x}\right)dx \end{aligned}

Why?

$\frac{y}{y + 2}$ is not a standard form. Write the top as $(y + 2) - 2$ ⁠:

\frac{y}{y + 2} = \frac{(y + 2) - 2}{y + 2}

That is $1 - \frac{2}{y + 2}$ ⁠. Similarly $\frac{x + 2}{x} = 1 + \frac{2}{x}$ ⁠.

\begin{aligned} \frac{(y + 2) - 2}{y + 2} &= \frac{y + 2}{y + 2} - \frac{2}{y + 2} \\ &= 1 - \frac{2}{y + 2} \\[4pt] \frac{x + 2}{x} &= \frac{x}{x} + \frac{2}{x} \\ &= 1 + \frac{2}{x} \end{aligned}

\begin{aligned} &y - 2\log|y + 2| \\ &= x + 2\log|x| + C \end{aligned}

Why?

$\int\frac{dy}{y + 2} = \log|y + 2|$ and $\int\frac{dx}{x} = \log|x|$ ⁠.

\begin{aligned} &\int 1\,dy - 2\int\frac{dy}{y + 2} \\ &= y - 2\log|y + 2| \\[4pt] &\int 1\,dx + 2\int\frac{dx}{x} \\ &= x + 2\log|x| \end{aligned}

One constant $C$ is added on the right.

Given $y(1) = -1$ ⁠: putting $x = 1$ ⁠, $y = -1$ ⁠,

\begin{aligned} -1 - 2\log 1 &= 1 + 2\log 1 + C \\ \Rightarrow C &= -2 \end{aligned}

Why?

$y(1) = -1$ means $y = -1$ when $x = 1$ ⁠. Then $|y + 2| = |1| = 1$ and $|x| = 1$ ⁠.

$\log 1 = 0$ ⁠, so $-1 = 1 + C$ ⁠, giving $C = -2$ ⁠.

\begin{aligned} \text{Left} &= -1 - 2\log|-1 + 2| \\ &= -1 - 2\log 1 \\ &= -1 - 2(0) = -1 \\[4pt] \text{Right} &= 1 + 2\log|1| + C \\ &= 1 + 2(0) + C = 1 + C \\[4pt] -1 &= 1 + C \\ C &= -1 - 1 = -2 \end{aligned}

$\therefore$ The particular solution is $y - 2\log|y + 2|$ $= x + 2\log|x| - 2$ ⁠.

Where this comes from

Chapter 9, Differential Equations: Homogeneous and variables-separable equations.

Question type: (a) Homogeneous equations; (b) Variables separable + A particular solution (combined)

How to spot it: click to see

(a) Type 6. Homogeneous equations: put $y = vx$

Spot it: every term has the same degree, like $x^2$ ⁠, $xy$ and $y^2$ ⁠, or you see $\frac{y}{x}$ inside sin, cos or log. You cannot separate $x$ and $y$ ⁠.

Here: Dividing by $x^2$ gives $1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2$ ⁠. Put $y = vx$ ⁠: the $v$ cancels, leaving $\frac{dv}{1 + v^2} = \frac{dx}{x}$ ⁠, a $\tan^{-1}$ ⁠.

(b) 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: The right side $(x + 2)(y + 2)$ is a part in $x$ times a part in $y$ ⁠. Separate: $\frac{y}{y + 2}\,dy = \frac{x + 2}{x}\,dx$ ⁠, and write $\frac{y}{y + 2} = 1 - \frac{2}{y + 2}$ before integrating.

(b) Type 4. A particular solution: solve, then find $C$

Spot it: a condition such as " $y = 1$ when $x = 0$ ", or "the curve passes through a point".

Here: Then $y(1) = -1$ ⁠: put $x = 1$ ⁠, $y = -1$ to get $-1 = 1 + C$ ⁠, so $C = -2$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

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Differential Equations in past papers: question 3 of 31 · see them all

Q29: Solve x2dydx=x2+xy+y2x^2\frac{dy}{dx} = x^2 + xy + y^2x2dxdy​=x2+xy+y2⁠, or xydydx=(x+2)(y+2)xy\frac{dy}{dx} = (x + 2)(y + 2)xydxdy​=(x+2)(y+2) with y(1)=−1y(1) = -1y(1)=−1

Q29: Solve $x^2\frac{dy}{dx} = x^2 + xy + y^2$ ⁠, or $xy\frac{dy}{dx} = (x + 2)(y + 2)$ with $y(1) = -1$