CBSE Sample Paper 2026-27 · Section C · 3 marks

Q29: Particular solution of $x\,dy + (y - x^2e^x)\,dx = 0$ ⁠, or of $(x^2 + y^2)\,dy = xy\,dx$

Sample 2026-27Ch 9 · Differential Equations

The question

(a) Find the particular solution of the differential equation

$x\,dy + (y - x^2e^x)\,dx = 0$ ⁠, given that y (1) =0

(b) Find the particular solution of the differential equation $(x^2 + y^2)\,dy = xy\,dx$ ⁠, given that $y(0) = 1$

Exam answer

(a)

Dividing by $dx$ ⁠,

x\frac{dy}{dx} + y - x^2e^x = 0

Dividing by $x$ ⁠, the equation is linear:

\frac{dy}{dx} + \frac1x\,y = xe^x

Why?

Dividing by $dx$ turns $dy$ into $\frac{dy}{dx}$ ⁠. Dividing by $x$ leaves $\frac{dy}{dx}$ on its own, and $\frac{x^2e^x}{x} = xe^x$ moves to the right.

This is the linear form

\frac{dy}{dx} + Py = Q

with $P = \frac1x$ and $Q = xe^x$ ⁠, both functions of $x$ only.

\text{I.F.} = e^{\int \frac1x\,dx} = e^{\log x} = x

Why?

The integrating factor is $e^{\int P\,dx}$ ⁠. Here $\int\frac1x\,dx = \log x$ (we may take $x > 0$ ⁠, since the given point has $x = 1$ ⁠).

Then $e^{\log x} = x$ ⁠, because $e$ and $\log$ undo each other.

The solution is

y \cdot x = \int x^2e^x\,dx

Why?

The solution of a linear equation is

y\cdot\text{I.F.} = \int Q\cdot\text{I.F.}\,dx + C

Here $Q\cdot\text{I.F.} = xe^x\cdot x = x^2e^x$ ⁠. The constant $C$ is added when the integral is done.

Using by parts,

yx = x^2e^x - \int 2xe^x\,dx

Using by parts again,

yx = x^2e^x - 2(xe^x - e^x) + C

Why?

By parts, with the algebraic factor as the first function and $e^x$ as the second:

\begin{aligned} \int x^2e^x\,dx &= x^2e^x \\ &\quad - \int 2x\,e^x\,dx \end{aligned}

Here the first function is $x^2$ ⁠, with derivative $2x$ ⁠, and $\int e^x\,dx = e^x$ ⁠.

The power of $x$ drops by one each time, so do it once more:

\int xe^x\,dx = xe^x - e^x

This time the first function is $x$ ⁠, with derivative 1:

\begin{aligned} \int xe^x\,dx &= x\,e^x - \int 1\cdot e^x\,dx \\ &= xe^x - e^x \\[4pt] \int 2xe^x\,dx &= 2(xe^x - e^x) \end{aligned}

Watch the signs: $-2(xe^x - e^x) = -2xe^x + 2e^x$ ⁠.

Given $y(1) = 0$ ⁠: $0 = e - 2(e - e) + C$ ⁠, so $C = -e$ ⁠.

Why?

Put $x = 1$ ⁠, $y = 0$ ⁠. The left side $yx$ is 0. On the right, $x^2e^x = e$ and $xe^x - e^x = e - e = 0$ ⁠. So $0 = e + C$ ⁠.

\begin{aligned} (0)(1) &= (1)^2e^1 \\ &\quad - 2\big((1)e^1 - e^1\big) + C \\ 0 &= e - 2(e - e) + C \\ 0 &= e - 2(0) + C \\ 0 &= e + C \\ C &= -e \end{aligned}

So $xy = x^2e^x - 2xe^x + 2e^x - e$ ⁠.

Why?

Put $C = -e$ and open the bracket:

\begin{aligned} xy &= x^2e^x - 2(xe^x - e^x) - e \\ &= x^2e^x - 2xe^x + 2e^x - e \end{aligned}

Then take out the common factor $e^x$ from the first three terms:

\begin{aligned} &x^2e^x - 2xe^x + 2e^x \\ &= e^x(x^2 - 2x + 2) \end{aligned}

$\therefore$ Particular solution: $xy = e^x(x^2 - 2x + 2) - e$

(b)

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

This is a homogeneous differential equation. Let $y = vx$ ⁠. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$ ⁠.

Why?

The top $xy$ and the bottom $x^2 + y^2$ both have degree 2 in every term. So the right side depends only on $\frac{y}{x}$ ⁠: the equation is homogeneous, and $y = vx$ is the standard substitution.

$\frac{dy}{dx} = v + x\frac{dv}{dx}$ comes from the product rule on $vx$ ⁠.

\begin{aligned} v + x\frac{dv}{dx} &= \frac{v}{1 + v^2} \\ \Rightarrow x\frac{dv}{dx} &= \frac{v - v - v^3}{1 + v^2} \\ &= \frac{-v^3}{1 + v^2} \end{aligned}

Why?

Put $y = vx$ on the right and cancel $x^2$ ⁠:

\frac{x\cdot vx}{x^2 + v^2x^2} = \frac{v}{1 + v^2}

\begin{aligned} \frac{x\cdot vx}{x^2 + v^2x^2} &= \frac{vx^2}{x^2(1 + v^2)} \\ &= \frac{v}{1 + v^2} \end{aligned}

Then subtract $v$ over the common denominator: $v = \frac{v + v^3}{1 + v^2}$ ⁠.

\begin{aligned} x\frac{dv}{dx} &= \frac{v}{1 + v^2} - v \\ &= \frac{v}{1 + v^2} - \frac{v(1 + v^2)}{1 + v^2} \\ &= \frac{v - v - v^3}{1 + v^2} \\ &= \frac{-v^3}{1 + v^2} \end{aligned}

Separating the variables,

\int \frac{1 + v^2}{v^3}\,dv = -\int \frac{dx}{x}

Why?

Put all the $v$ terms with $dv$ and all the $x$ terms with $dx$ ⁠:

Multiply both sides by $\frac{1 + v^2}{-v^3}$ and by $\frac{dx}{x}$ ⁠:

\frac{1 + v^2}{-v^3}\,dv = \frac{dx}{x}

Then move the minus sign to the right and integrate both sides.

\begin{aligned} &\int\left(\frac{1}{v^3} + \frac1v\right)dv = -\int\frac{dx}{x} \\ \Rightarrow\ &-\frac{1}{2v^2} + \log|v| \\ &\quad = -\log|x| + \log C \end{aligned}

Why?

Split the fraction: $\frac{1 + v^2}{v^3} = \frac{1}{v^3} + \frac{1}{v}$ ⁠.

\frac{1 + v^2}{v^3} = \frac{1}{v^3} + \frac{v^2}{v^3} = \frac{1}{v^3} + \frac{1}{v}

Also $\int\frac1v\,dv = \log|v|$ and $-\int\frac{dx}{x} = -\log|x|$ ⁠.

Power rule with $n = -3$ ⁠:

\int v^{-3}\,dv = \frac{v^{-2}}{-2} = -\frac{1}{2v^2}

Writing the constant as $\log C$ makes the logs easy to combine later.

Putting $v = \frac{y}{x}$ ⁠,

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

\log\left|\frac{y}{C}\right| = \frac{x^2}{2y^2}

Why?

With $v = \frac{y}{x}$ ⁠: $\frac{1}{2v^2} = \frac{x^2}{2y^2}$ and $\log|v| = \log|y| - \log|x|$ ⁠.

The $-\log|x|$ cancels from both sides. Then $\log|y| - \log C = \log\left|\frac{y}{C}\right|$ ⁠.

\begin{aligned} \frac{1}{2v^2} &= \frac{1}{2\left(\frac{y}{x}\right)^2} = \frac{1}{2\cdot\frac{y^2}{x^2}} \\ &= \frac{x^2}{2y^2} \\[4pt] -\frac{x^2}{2y^2} + \log|y| &= \log C \\ \log|y| - \log C &= \frac{x^2}{2y^2} \\ \log\left|\frac{y}{C}\right| &= \frac{x^2}{2y^2} \end{aligned}

When $x = 0$ ⁠, $y = 1$ ⁠: $\log\frac1C = 0$ ⁠, so $C = 1$ ⁠.

So $\log y = \frac{x^2}{2y^2}$ ⁠.

Why?

At $x = 0$ the right side is 0. The only number whose log is 0 is 1, so $\frac1C = 1$ ⁠.

\begin{aligned} \log\left|\frac{1}{C}\right| &= \frac{(0)^2}{2(1)^2} = \frac{0}{2} = 0 \\ \frac{1}{C} &= e^0 = 1 \\ C &= 1 \end{aligned}

So $\log\left|\frac{y}{1}\right| = \log|y| = \frac{x^2}{2y^2}$ ⁠.

Near the point $(0, 1)$ ⁠, $y$ is positive, so $|y| = y$ ⁠. Then $\log y = k$ means $y = e^k$ ⁠.

$\therefore$ The particular solution is $y = e^{\dfrac{x^2}{2y^2}}$

Where this comes from

Chapter 9, Differential Equations: Linear and homogeneous differential equations.

Question type: (a) Linear equations + Integration by parts + A particular solution (combined); (b) Homogeneous equations + A particular solution (combined)

How to spot it: click to see

(a) Ch 9, type 7. Linear equations: the integrating factor

Spot it: $\frac{dy}{dx} + Py = Q$ ⁠, with $P$ and $Q$ in $x$ only, and $y$ only to power 1.

Here: Divide by $x\,dx$ ⁠: $\frac{dy}{dx} + \frac1x y = xe^x$ ⁠. So $P = \frac1x$ ⁠, $\text{I.F.} = x$ ⁠, and $yx = \int x^2e^x\,dx$ ⁠.

(a) 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: $\int x^2e^x\,dx$ is a power times $e^x$ ⁠: integrate by parts twice, with $x^2$ and then $x$ as the first function.

(a) Ch 9, 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 $y(1) = 0$ gives $0 = e - 2(e - e) + C$ ⁠, so $C = -e$ ⁠.

(b) Ch 9, 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: $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$ has every term of degree 2. Put $y = vx$ ⁠: $x\frac{dv}{dx} = \frac{-v^3}{1 + v^2}$ ⁠, so separate $\frac{1 + v^2}{v^3}\,dv = -\frac{dx}{x}$ ⁠.

(b) Ch 9, 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 $y(0) = 1$ gives $\log\frac1C = 0$ ⁠, so $C = 1$ and $y = e^{\frac{x^2}{2y^2}}$ ⁠.

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Q29: Particular solution of x dy+(y−x2ex) dx=0x\,dy + (y - x^2e^x)\,dx = 0xdy+(y−x2ex)dx=0⁠, or of (x2+y2) dy=xy dx(x^2 + y^2)\,dy = xy\,dx(x2+y2)dy=xydx

Q29: Particular solution of $x\,dy + (y - x^2e^x)\,dx = 0$ ⁠, or of $(x^2 + y^2)\,dy = xy\,dx$