CBSE Sample Paper 2023-24 · Section C · 3 marks

Q29: Solve $ye^{\frac{x}{y}}\,dx = \left(xe^{\frac{x}{y}} + y^2\right)dy$ ⁠, or $(\cos^2 x)\frac{dy}{dx} + y = \tan x$

Sample 2023-24Ch 9 · Differential Equations

The question

(a) Solve the differential equation: $ye^{\frac{x}{y}}\,dx = \left(xe^{\frac{x}{y}} + y^2\right)dy$ ⁠, $(y \ne 0)$ ⁠.

(b) Solve the differential equation: $(\cos^2 x)\dfrac{dy}{dx} + y = \tan x$ ⁠; $\left(0 \le x < \dfrac{\pi}{2}\right)$ ⁠.

Exam answer

(a) We have

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

Why?

Divide both sides by $y\,e^{\frac{x}{y}}\,dy$ ⁠. The right side is then built around $\frac{x}{y}$ ⁠, so we treat $x$ as the unknown function of $y$ and work with $\frac{dx}{dy}$ ⁠.

On the left $ye^{\frac{x}{y}}$ cancels, leaving $\frac{dx}{dy}$ ⁠. On the right only $dy$ cancels.

\Rightarrow \frac{dx}{dy} = \frac{x}{y} + \frac{y}{e^{\frac{x}{y}}}\quad\ldots(\text{i})

Why?

Split the fraction into two:

\frac{xe^{\frac{x}{y}}}{y\,e^{\frac{x}{y}}} = \frac{x}{y},\quad \frac{y^2}{y\,e^{\frac{x}{y}}} = \frac{y}{e^{\frac{x}{y}}}

Let $x = vy \Rightarrow \dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$ ⁠.

Why?

The substitution $x = vy$ turns $\frac{x}{y}$ into the single letter $v$ ⁠.

Both $v$ and $y$ vary, so differentiate $vy$ with respect to $y$ by the product rule: $v \cdot 1 + y\frac{dv}{dy}$ ⁠.

Watch out: $\frac{dx}{dy}$ is not just $v$ ⁠.

So equation (i) becomes

v + y\frac{dv}{dy} = v + \frac{y}{e^v}

\Rightarrow y\frac{dv}{dy} = \frac{y}{e^v}

\Rightarrow e^v\,dv = dy

Why?

In (i), put $v + y\frac{dv}{dy}$ for $\frac{dx}{dy}$ and $v$ for $\frac{x}{y}$ ⁠. So $e^{\frac{x}{y}} = e^v$ ⁠.

$v$ is on both sides, so it cancels. Then divide by $y$ (allowed, as $y \ne 0$ ⁠) to get $\frac{dv}{dy} = \frac{1}{e^v}$ ⁠.

Multiply by $e^v\,dy$ to separate the variables: everything in $v$ on one side, everything in $y$ on the other.

\begin{aligned} v + y\frac{dv}{dy} &= v + \frac{y}{e^v} \\ y\frac{dv}{dy} &= \frac{y}{e^v} \quad (\text{subtract } v) \\ \frac{dv}{dy} &= \frac{1}{e^v} \quad (\text{divide by } y) \\ e^v\,dv &= dy \quad (\times\, e^v\,dy) \end{aligned}

On integrating, we get

\int e^v\,dv = \int dy \Rightarrow e^v = y + c

Why?

$\int e^v\,dv = e^v$ and $\int dy = y$ ⁠. One constant $c$ on one side is enough.

Then put $v = \frac{x}{y}$ back to return to $x$ and $y$ ⁠.

$\Rightarrow e^{\frac{x}{y}} = y + c$ ⁠, where $c$ is an arbitrary constant of integration.

(b) Dividing both the sides by $\cos^2 x$ ⁠, we get

\begin{aligned} &\frac{dy}{dx} + y\sec^2 x \\ &= \tan x\sec^2 x\quad\ldots(\text{i}) \end{aligned}

Why?

The standard linear form needs $\frac{dy}{dx}$ on its own, with coefficient 1, so divide every term by $\cos^2 x$ ⁠. This is allowed because $\cos x \ne 0$ for $0 \le x < \frac{\pi}{2}$ ⁠.

$\frac{1}{\cos^2 x} = \sec^2 x$ ⁠, and $\frac{\tan x}{\cos^2 x} = \tan x\sec^2 x$ ⁠.

Comparing with $\dfrac{dy}{dx} + Py = Q$ ⁠: $P = \sec^2 x$ ⁠, $Q = \tan x\sec^2 x$ ⁠.

\text{I.F.} = e^{\int\sec^2 x\,dx} = e^{\tan x}

Why?

The integrating factor of $\frac{dy}{dx} + Py = Q$ is

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

Here $\int\sec^2 x\,dx = \tan x$ ⁠, because $\frac{d}{dx}\tan x = \sec^2 x$ ⁠. No constant is needed in the I.F.

On multiplying equation (i) by $e^{\tan x}$ ⁠, we get

\frac{d}{dx}\left(ye^{\tan x}\right) = e^{\tan x}\tan x\sec^2 x

Why?

The I.F. is chosen so that the left side becomes the derivative of a product. By the product and chain rules,

\begin{aligned} &\frac{d}{dx}\left(ye^{\tan x}\right) \\ &= e^{\tan x}\frac{dy}{dx} + ye^{\tan x}\sec^2 x \end{aligned}

which is exactly the left side of (i) times $e^{\tan x}$ ⁠.

On integrating, with $t = \tan x$ ⁠, so that $dt = \sec^2 x\,dx$ ⁠,

\begin{aligned} ye^{\tan x} &= \int te^t\,dt \\ &= te^t - \int e^t\,dt \\ &= te^t - e^t + c \\ &= (\tan x - 1)e^{\tan x} + c \end{aligned}

Why?

With $t = \tan x$ ⁠, the right side $e^{\tan x}\tan x\sec^2 x\,dx$ becomes $te^t\,dt$ ⁠.

Integrate by parts with $t$ as the first function and $e^t$ as the second:

(Inside the integral: $\tan x$ becomes $t$ ⁠, $e^{\tan x}$ becomes $e^t$ ⁠, and $\sec^2 x\,dx$ becomes $dt$ ⁠.)

\begin{aligned} u &= t, & \frac{du}{dt} &= 1 \\ \int e^t\,dt &= e^t \end{aligned}

\int te^t\,dt = te^t - \int 1\cdot e^t\,dt

\begin{aligned} &te^t - \int e^t\,dt \\ &= te^t - e^t + c \\ &= (t - 1)e^t + c \end{aligned}

(The last line takes out $e^t$ ⁠.)

Put $t = \tan x$ back, then divide both sides by $e^{\tan x}$ ⁠: $\frac{c}{e^{\tan x}} = ce^{-\tan x}$ ⁠.

\begin{aligned} ye^{\tan x} &= (\tan x - 1)e^{\tan x} + c \\ y &= \frac{(\tan x - 1)e^{\tan x}}{e^{\tan x}} + \frac{c}{e^{\tan x}} \\ &= \tan x - 1 + ce^{-\tan x} \end{aligned}

$\therefore\ y = \tan x - 1 + ce^{-\tan x}$ ⁠, where $c$ is an arbitrary constant of integration.

Where this comes from

Chapter 9, Differential Equations: Differential equations: x = vy; linear equation.

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

How to spot it: click to see

(a) 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 turn the $y = vx$ idea round: write $\frac{dx}{dy}$ and put $x = vy$ ⁠. The $v$ terms cancel, leaving $e^v\,dv = dy$ ⁠.

(b) 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 $\cos^2x$ so that $\frac{dy}{dx}$ stands alone: $P = \sec^2x$ ⁠, $Q = \tan x\sec^2x$ ⁠, and $\text{I.F.} = e^{\tan x}$ ⁠.

(b) 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: With $t = \tan x$ ⁠, the right side becomes $\int te^t\,dt$ ⁠: a product of $t$ and $e^t$ ⁠, so integrate by parts, with $t$ as the first function.

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Differential Equations in past papers: question 23 of 31 · see them all

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Q29: Solve $ye^{\frac{x}{y}}\,dx = \left(xe^{\frac{x}{y}} + y^2\right)dy$ ⁠, or $(\cos^2 x)\frac{dy}{dx} + y = \tan x$