CBSE Sample Paper 2025-26 · Section D · 5 marks

Q34: Solve $y + \frac{d}{dx}(xy) = x(\sin x + x)$ ⁠, or the particular solution of $2ye^{x/y}dx + (y - 2xe^{x/y})dy = 0$

Sample 2025-26Ch 9 · Differential Equations

The question

(a) Solve the differential equation: $y + \dfrac{d}{dx}(xy) = x(\sin x + x)$

(b) Find the particular solution of the differential equation:

2ye^{x/y}dx + (y - 2xe^{x/y})\,dy = 0

given that $y(0) = 1$

Exam answer

(a)

\begin{aligned} &y + \frac{d}{dx}(xy) = x(\sin x + x) \\ \Rightarrow\ &y + \left(x\frac{dy}{dx} + y\right) = x(\sin x + x) \\ \Rightarrow\ &2y + x\frac{dy}{dx} = x(\sin x + x) \\ \Rightarrow\ &\frac{dy}{dx} + \frac{2}{x}y = \sin x + x \end{aligned}

Why?

$\frac{d}{dx}(xy)$ needs the product rule, since both $x$ and $y$ change:

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

Then collect $y + y = 2y$ ⁠, and divide every term by $x$ so that $\frac{dy}{dx}$ stands alone.

\begin{aligned} &\frac{x}{x}\frac{dy}{dx} + \frac{2y}{x} = \frac{x(\sin x + x)}{x} \\ &\frac{dy}{dx} + \frac{2}{x}y = \sin x + x \end{aligned}

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$ ⁠, with $P = \frac{2}{x}$ ⁠, $Q = \sin x + x$ ⁠.

Why?

It is linear because $y$ and $\frac{dy}{dx}$ appear only to the first power, and $P$ and $Q$ are functions of $x$ alone.

\text{I.F.} = e^{\int\frac{2}{x}dx} = e^{2\log x} = e^{\log x^2} = x^2

Why?

The integrating factor is $e^{\int P\,dx}$ ⁠. Here $\int\frac{2}{x}\,dx = 2\log x$ ⁠, and $2\log x = \log x^2$ ⁠.

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

The solution is $y\cdot\text{I.F.} = \displaystyle\int Q\cdot\text{I.F.}\,dx$ ⁠:

\begin{aligned} yx^2 &= \int(\sin x + x)x^2\,dx \\ &= \int x^2\sin x\,dx + \int x^3\,dx \end{aligned}

Why?

Multiplying the equation by the I.F. makes the left side an exact derivative:

\begin{aligned} &x^2\frac{dy}{dx} + 2xy \\ &= \frac{d}{dx}(x^2y) \end{aligned}

So integrating both sides gives the solution formula. Then multiply out: $(\sin x + x)x^2 = x^2\sin x + x^3$ ⁠.

Integrating by parts ( $x^2$ as the first function),

\begin{aligned} &\int x^2\sin x\,dx \\ &= x^2(-\cos x) - \int 2x(-\cos x)\,dx \end{aligned}

Why?

By parts:

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

Take $u = x^2$ ⁠, because differentiating lowers its power: $u' = 2x$ ⁠. Take $v = \sin x$ ⁠, with $\int\sin x\,dx = -\cos x$ ⁠.

\begin{aligned} yx^2 &= -x^2\cos x + 2\int x\cos x\,dx \\ &\quad + \frac{x^4}{4} + C \end{aligned}

Why?

The two minus signs make a plus: $-\int 2x(-\cos x)\,dx = 2\int x\cos x\,dx$ ⁠.

By the power rule, $\int x^3\,dx = \frac{x^4}{4}$ ⁠. One constant $C$ is enough for the whole right side.

Again by parts ( $x$ as the first function),

\begin{aligned} &\int x\cos x\,dx \\ &= x\sin x - \int 1\cdot\sin x\,dx \\ &= x\sin x + \cos x \end{aligned}

Why?

Now $u = x$ ⁠, $u' = 1$ ⁠, and $v = \cos x$ with $\int\cos x\,dx = \sin x$ ⁠.

Then $\int\sin x\,dx = -\cos x$ ⁠, so subtracting it gives $+\cos x$ ⁠.

\begin{aligned} &x\sin x - \int\sin x\,dx \\ &= x\sin x - (-\cos x) \\ &= x\sin x + \cos x \end{aligned}

(The constant is already in $C$ ⁠.)

\begin{aligned} yx^2 &= -x^2\cos x \\ &\quad + 2(x\sin x + \cos x) \\ &\quad + \frac{x^4}{4} + C \end{aligned}

Why?

This is the general solution: it has one arbitrary constant $C$ ⁠, as a first-order equation should. Dividing by $x^2$ would give $y$ explicitly, but this form is accepted.

This is the required (general) solution.

(b)

2ye^{x/y}\,dx = (2xe^{x/y} - y)\,dy

Why?

Move the $dy$ term to the other side and change its sign: $-(y - 2xe^{x/y}) = 2xe^{x/y} - y$ ⁠.

\begin{aligned} \frac{dx}{dy} &= \frac{2xe^{x/y} - y}{2ye^{x/y}} \\ &= \frac{2\frac{x}{y}e^{x/y} - 1}{2e^{x/y}} \end{aligned}

Why?

The equation contains $e^{x/y}$ ⁠, so it is easier to treat $x$ as a function of $y$ and write $\frac{dx}{dy}$ ⁠.

Dividing the top and the bottom by $y$ shows that $\frac{dx}{dy}$ depends only on $\frac{x}{y}$ ⁠. That is what homogeneous means here.

\begin{aligned} \text{top} \div y&: \frac{2xe^{x/y}}{y} - \frac{y}{y} \\ &= 2\frac{x}{y}e^{x/y} - 1 \\[4pt] \text{bottom} \div y&: \frac{2ye^{x/y}}{y} = 2e^{x/y} \end{aligned}

It is a homogeneous differential equation. Let $x = vy \Rightarrow \frac{dx}{dy} = v + y\frac{dv}{dy}$ ⁠.

Why?

With $x = vy$ ⁠, the ratio $\frac{x}{y}$ is just $v$ ⁠. Both $v$ and $y$ vary, so the product rule gives

\frac{d}{dy}(vy) = v + y\frac{dv}{dy}

\begin{aligned} v + y\frac{dv}{dy} &= \frac{2ve^v - 1}{2e^v} \\ \Rightarrow y\frac{dv}{dy} &= \frac{2ve^v - 1}{2e^v} - v \\ &= \frac{2ve^v - 1 - 2ve^v}{2e^v} \\ &= \frac{-1}{2e^v} \\ \Rightarrow 2e^v\,dv &= -\frac{dy}{y} \end{aligned}

Why?

First put $\frac{x}{y} = v$ into the right side, then subtract $v$ from both sides.

Write $v$ as $\frac{2ve^v}{2e^v}$ to subtract it with a common denominator. The $2ve^v$ terms cancel.

\begin{aligned} &\frac{2ve^v - 1}{2e^v} - \frac{2ve^v}{2e^v} \\ &= \frac{2ve^v - 1 - 2ve^v}{2e^v} = \frac{-1}{2e^v} \end{aligned}

Then separate the variables, $v$ with $dv$ and $y$ with $dy$ ⁠: multiply both sides by $\frac{2e^v}{y}\,dy$ ⁠.

\begin{aligned} y\frac{dv}{dy} \cdot \frac{2e^v}{y}\,dy &= 2e^v\,dv \\ \frac{-1}{2e^v} \cdot \frac{2e^v}{y}\,dy &= -\frac{dy}{y} \end{aligned}

Integrating both sides,

\begin{aligned} \int 2e^v\,dv &= -\int\frac{dy}{y} \\ \Rightarrow 2e^v &= -\log|y| + C \\ \Rightarrow 2e^{x/y} + \log|y| &= C \end{aligned}

Why?

Standard integrals: $\int e^v\,dv = e^v$ and $\int\frac{dy}{y} = \log|y|$ ⁠.

Then put $v = \frac{x}{y}$ back, and move $\log|y|$ to the left.

When $x = 0$ ⁠, $y = 1$ ⁠: $2e^0 + \log 1 = C$ ⁠, so $C = 2$ ⁠.

Why?

$y(0) = 1$ means $y = 1$ when $x = 0$ ⁠. Then $\frac{x}{y} = 0$ ⁠, $e^0 = 1$ and $\log 1 = 0$ ⁠, so $C = 2 + 0 = 2$ ⁠.

$\therefore$ The required solution is $2e^{x/y} + \log|y| = 2$ ⁠.

Where this comes from

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

Question type: (a) Linear equations + Integration by parts (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: Expand $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$ ⁠, then divide by $x$ ⁠: $\frac{dy}{dx} + \frac{2}{x}y = \sin x + x$ ⁠. So $\text{I.F.} = x^2$ ⁠.

(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: The right side needs $\int x^2\sin x\,dx$ ⁠: a power times a trig function, so integrate by parts twice, with $x^2$ and then $x$ as the first function.

(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: $\frac{x}{y}$ sits inside $e^{x/y}$ ⁠, so write $\frac{dx}{dy} = \frac{2\frac{x}{y}e^{x/y} - 1}{2e^{x/y}}$ and put $x = vy$ ⁠. The $v$ terms cancel, leaving $2e^v\,dv = -\frac{dy}{y}$ ⁠.

(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 $2e^0 + \log 1 = C$ ⁠, so $C = 2$ ⁠.

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Differential Equations in past papers: question 18 of 31 · see them all

Q34: Solve y+ddx(xy)=x(sin⁡x+x)y + \frac{d}{dx}(xy) = x(\sin x + x)y+dxd​(xy)=x(sinx+x)⁠, or the particular solution of 2yex/ydx+(y−2xex/y)dy=02ye^{x/y}dx + (y - 2xe^{x/y})dy = 02yex/ydx+(y−2xex/y)dy=0

Q34: Solve $y + \frac{d}{dx}(xy) = x(\sin x + x)$ ⁠, or the particular solution of $2ye^{x/y}dx + (y - 2xe^{x/y})dy = 0$