CBSE Board Examination 2023 · Section C · 3 marks

Q28: Solve $\frac{dy}{dx} + \sec^2x\cdot y = \tan x\cdot\sec^2x$ ⁠, $y(0) = 0$ ⁠, or $x\,dy - y\,dx - \sqrt{x^2 + y^2}\,dx = 0$

Board 2023Ch 9 · Differential Equations

The question

(a) Find the particular solution of the differential equation $\dfrac{dy}{dx} + \sec^2x\,.\,y = \tan x\,.\,\sec^2x$ ⁠, given that $y(0) = 0$ ⁠.

(b) Solve the differential equation given by

x\,dy - y\,dx - \sqrt{x^2 + y^2}\,dx = 0.

Exam answer

(a) This is linear, of the form $\frac{dy}{dx} + Py = Q$ with $P = \sec^2x$ ⁠, $Q = \tan x\sec^2x$ ⁠.

Why?

A linear equation has $y$ and $\frac{dy}{dx}$ only to the first power, and $P$ and $Q$ depend on $x$ alone. $P$ is the coefficient of $y$ ⁠, and $Q$ is the right side.

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

Why?

The integrating factor is

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

and $\int\sec^2x\,dx = \tan x$ ⁠, because the derivative of $\tan x$ is $\sec^2x$ ⁠.

Solution is

\begin{aligned} &y\,e^{\tan x} \\ &= \int\tan x\sec^2x\,e^{\tan x}\,dx + C \end{aligned}

Why?

For a linear equation the solution is

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

This works because multiplying the equation by the I.F. turns its left side into the derivative of $y\,e^{\tan x}$ ⁠.

Let $\tan x = t$ ⁠, $\sec^2x\,dx = dt$ ⁠.

\begin{aligned} &\int e^{\tan x}\tan x\sec^2x\,dx \\ &= \int e^t\,t\,dt = te^t - \int e^t\,dt \\ &= e^t(t - 1) \end{aligned}

Why?

$\sec^2x\,dx$ is exactly the differential of $\tan x$ ⁠, so $t = \tan x$ leaves $\int t\,e^t\,dt$ ⁠.

That needs integration by parts, with $t$ as the first function:

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

Here the first function is $t$ ⁠, with derivative 1, and the second is $e^t$ ⁠, with integral $e^t$ ⁠.

Since $\int e^t\,dt = e^t$ ⁠, this is $te^t - e^t = e^t(t - 1)$ ⁠. Then put back $t = \tan x$ ⁠.

\begin{aligned} te^t - e^t &= e^t(t) - e^t(1) \\ &= e^t(t - 1) \\ &= e^{\tan x}(\tan x - 1) \end{aligned}

\therefore\ y\,e^{\tan x} = e^{\tan x}(\tan x - 1) + C

$y(0) = 0$ gives $0 = e^0(0 - 1) + C$ ⁠, so $C = 1$ ⁠.

Why?

$y(0) = 0$ means $y = 0$ when $x = 0$ ⁠. With $\tan 0 = 0$ and $e^0 = 1$ ⁠, the equation becomes $0 = -1 + C$ ⁠.

\begin{aligned} 0 \cdot e^{\tan 0} &= e^{\tan 0}(\tan 0 - 1) + C \\ 0 \cdot e^0 &= e^0(0 - 1) + C \\ 0 &= 1(-1) + C \\ 0 &= -1 + C \\ C &= 1 \end{aligned}

Particular solution is $y\,e^{\tan x} = e^{\tan x}(\tan x - 1) + 1$ ⁠, or $y = \tan x - 1 + e^{-\tan x}$ ⁠.

Why?

For the second form, divide both sides by $e^{\tan x}$ ⁠, using

\frac{1}{e^{\tan x}} = e^{-\tan x}

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

(b) $x\,dy = \left(y + \sqrt{x^2 + y^2}\right)dx$

The equation can be written as (taking $x > 0$ ⁠)

\frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2}\quad\ldots(\text{i})

Why?

Divide by $x\,dx$ ⁠:

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

For $x > 0$ ⁠, $x = \sqrt{x^2}$ ⁠, so the root term is

\begin{aligned} \frac{\sqrt{x^2 + y^2}}{x} &= \sqrt{\frac{x^2 + y^2}{x^2}} \\ &= \sqrt{1 + \left(\frac{y}{x}\right)^2} \end{aligned}

It is homogeneous. Let $y = vx$ ⁠, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$ ⁠. Substituting in (i),

Why?

The right side of (i) depends only on $\frac{y}{x}$ ⁠, so the equation is homogeneous. The standard substitution for such an equation is $y = vx$ ⁠.

By the product rule, $\frac{d}{dx}(vx) = v + x\frac{dv}{dx}$ ⁠.

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

Why?

On the right of (i), $\frac{y}{x} = \frac{vx}{x} = v$ ⁠.

$v$ cancels from both sides, which leaves $x\frac{dv}{dx} = \sqrt{1 + v^2}$ ⁠.

Then separate the variables: every $v$ goes with $dv$ ⁠, and every $x$ goes with $dx$ ⁠.

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

(Multiply by $dx$ ⁠, then divide by $x\sqrt{1 + v^2}$ ⁠.)

Integrating both sides,

\begin{aligned} &\log\left|v + \sqrt{1 + v^2}\right| \\ &= \log|x| + \log C \\ \Rightarrow\ &v + \sqrt{1 + v^2} = Cx \\ \Rightarrow\ &\frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = Cx \end{aligned}

Why?

Standard integral, here with $a = 1$ ⁠:

\begin{aligned} &\int\frac{dv}{\sqrt{v^2 + a^2}} \\ &= \log\left|v + \sqrt{v^2 + a^2}\right| \end{aligned}

Writing the constant as $\log C$ lets the logs combine: $\log|x| + \log C = \log C|x|$ ⁠. Equal logs give equal numbers. As $x > 0$ and $v + \sqrt{1 + v^2} > 0$ ⁠, no modulus is needed.

Finally put back $v = \frac{y}{x}$ and multiply through by $x$ ⁠.

\begin{aligned} \sqrt{1 + v^2} &= \sqrt{1 + \frac{y^2}{x^2}} \\ &= \sqrt{\frac{x^2 + y^2}{x^2}} \\ &= \frac{\sqrt{x^2 + y^2}}{x} \end{aligned}

\begin{aligned} &x\left(\frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x}\right) = x(Cx) \\ &y + \sqrt{x^2 + y^2} = Cx^2 \end{aligned}

$\therefore\ y + \sqrt{x^2 + y^2} = Cx^2$ ⁠.

Where this comes from

Chapter 9, Differential Equations: Linear differential equation; homogeneous equation.

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

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: Here $P = \sec^2x$ and $Q = \tan x\sec^2x$ ⁠, both in $x$ only, so $\text{I.F.} = e^{\tan x}$ ⁠.

(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: With $t = \tan x$ ⁠, the right side $\int\tan x\sec^2x\,e^{\tan x}\,dx$ becomes $\int te^t\,dt$ ⁠: a product of $t$ and $e^t$ ⁠, so integrate by parts, with $t$ 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 $y(0) = 0$ gives $0 = e^0(0 - 1) + C$ ⁠, so $C = 1$ ⁠.

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

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Differential Equations in past papers: question 11 of 31 · see them all

Q28: Solve dydx+sec⁡2x⋅y=tan⁡x⋅sec⁡2x\frac{dy}{dx} + \sec^2x\cdot y = \tan x\cdot\sec^2xdxdy​+sec2x⋅y=tanx⋅sec2x⁠, y(0)=0y(0) = 0y(0)=0⁠, or x dy−y dx−x2+y2 dx=0x\,dy - y\,dx - \sqrt{x^2 + y^2}\,dx = 0xdy−ydx−x2+y2​dx=0

Q28: Solve $\frac{dy}{dx} + \sec^2x\cdot y = \tan x\cdot\sec^2x$ ⁠, $y(0) = 0$ ⁠, or $x\,dy - y\,dx - \sqrt{x^2 + y^2}\,dx = 0$