As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced..

**Q1.**If the integrating factor of the differential equation 𝑑𝑦/𝑑𝑥+ 𝑃(𝑥)𝑦 = 𝑄(𝑥) is 𝑥, then 𝑃(𝑥) is

Solution

Given, IF = 𝑥

Given, IF = 𝑥

∴ 𝑒
∫𝑃 𝑑𝑥 = 𝑥

⇒ ∫ 𝑃 𝑑𝑥 = log 𝑥

⇒ 𝑃 =
𝑑/𝑑𝑥 (log 𝑥) =
1/x

**Q2.**The differential equation of all circles which passes through the origin and whose centre lies on 𝑦- axis, is

Solution

Let 𝑥² + 𝑦² − 2𝑘𝑦 = 0

Let 𝑥² + 𝑦² − 2𝑘𝑦 = 0

⇒ 2𝑥 + 2𝑦
𝑑𝑦/𝑑𝑥
− 2𝑘
𝑑𝑦/𝑑𝑥
= 0

⇒ 𝑘 =
𝑘/(
𝑑𝑦/𝑑𝑥)
+ 𝑦

From Eq. (i),

𝑥² + 𝑦² − 2 (
𝑥 /(𝑑𝑦⁄𝑑𝑥)
+ 𝑦) 𝑦 = 0

⇒ (𝑥² − 𝑦² )
𝑑𝑦/𝑑𝑥
− 2𝑥𝑦 = 0

**Q3.**The differential equation obtained by eliminating arbitrary constants from 𝑦 = 𝑎e

^{bx}is

Solution

The given equation is 𝑦 = 𝑎e

The given equation is 𝑦 = 𝑎e

^{bx}⇒
𝑑𝑦/𝑑𝑥
= 𝑎𝑏e

^{bx}…(i)⇒
𝑑²𝑦/𝑑𝑥² = 𝑎𝑏
2e

^{bx}…(ii)⇒ 𝑎e

^{bx}𝑑 2𝑦 𝑑𝑥2 = 𝑎²𝑏² e^{bx}e^{bx}⇒ 𝑦
𝑑²𝑦/𝑑𝑥² = (
𝑑𝑦/𝑑𝑥)² [from eq. (ii)]

**Q4.**A continuously differential function ϕ(𝑥) in (0, 𝜋) satisfying 𝑦 ′ = 1 + 𝑦² ,𝑦(0) = 0 = 𝑦(𝜋), is

Solution

Given that, 𝑑𝑦/𝑑𝑥 = 1 + 𝑦² ⇒ 𝑑𝑦/(1 + 𝑦²) = 𝑑𝑥

Given that, 𝑑𝑦/𝑑𝑥 = 1 + 𝑦² ⇒ 𝑑𝑦/(1 + 𝑦²) = 𝑑𝑥

On integrating both sides, we get

∫
𝑑𝑦/(1 + 𝑦²) = ∫ 𝑑𝑥

⇒ tan−1 𝑦 = 𝑥 + 𝑐

At 𝑥 = 0, 𝑦 = 0, then 𝑐 = 0
At 𝑥 = 𝜋, 𝑦 = 0,

then tan−1 0 = 𝜋 + 𝑐 ⇒ 𝑐 = −𝜋

∴ tan−1 𝑦 = 𝑥 ⇒ 𝑦 = tan 𝑥 = ϕ(𝑥)

Therefore, solution becomes 𝑦 = tan 𝑥
But tan 𝑥 is not continuous function in (0, 𝜋)
So, ϕ(𝑥) is not possible in (0, 𝜋).

**Q5.**Solution of the differential equation 𝑥 𝑑𝑦 − 𝑦 𝑑𝑥 = 0 represents

Solution

We have, 𝑥 𝑑𝑦 − 𝑦 𝑑𝑥 = 0

We have, 𝑥 𝑑𝑦 − 𝑦 𝑑𝑥 = 0

⇒
𝑑𝑦/𝑦
−
𝑑𝑥/𝑥
= 0

⇒ log 𝑦 − log 𝑥 = log 𝐶 [On integrating]

⇒
𝑦/𝑥
= 𝐶 ⇒ 𝑦 = 𝐶 𝑥

Clearly, it represents a family of straight lines
passing through the origin

**Q7.**The differential equation whose solution is 𝐴𝑥² + 𝐵𝑦² = 1, where 𝐴 and 𝐵 are arbitrary constants, is

of

Solution

The given equation is 𝐴𝑥² + 𝐵𝑦² = 1

The given equation is 𝐴𝑥² + 𝐵𝑦² = 1

⇒ 2𝐴𝑥 + 2𝐵𝑦 𝑑𝑦/𝑑𝑥
= 0
…(i)

⇒ 2𝐴 + 2𝐵 {(
𝑑𝑦/𝑑𝑥)² + 𝑦
𝑑²𝑦/𝑑𝑥² } = 0
…(ii)

Eliminating A and B from Eqs. (i) and (ii),

we
get
𝑦
𝑑²𝑦/𝑑² + (
𝑑𝑦/𝑑𝑥)² −
𝑦/𝑥.𝑑𝑦/𝑑𝑥
= 0

Here, order =2, degree =1

**Q8.**The degree of the differential equation of all curves having normal of constant length 𝑐, is

Solution

We have,

We have,

Clearly, it is a differential equation of degree 2

**Q9.**The solution of the differential equation 𝑑𝑦/𝑑𝑥=𝑦/𝑥+ϕ (𝑦/𝑥)/ϕ′(𝑦/𝑥)is

Solution

Given equation is, 𝑑𝑦/𝑑𝑥 = 𝑦/𝑥 + ϕ( 𝑦/𝑥 )/ϕ′( 𝑦/𝑥 ) ...(i)

Given equation is, 𝑑𝑦/𝑑𝑥 = 𝑦/𝑥 + ϕ( 𝑦/𝑥 )/ϕ′( 𝑦/𝑥 ) ...(i)

Put 𝑦 = 𝑣𝑥 ⇒
𝑑𝑦/𝑑𝑥
= 𝑣 + 𝑥
𝑑𝑣/𝑑𝑥

Now, Eq. (i) becomes
𝑣 + 𝑥
𝑑𝑣/𝑑𝑥 = 𝑣 +
ϕ(𝑣)/ϕ′(𝑣)

⇒
ϕ′(𝑣)/ϕ(𝑣)
𝑑𝑣 =
𝑑𝑥/𝑥

On integrating both sides, we get

∫
ϕ′(𝑣)/ϕ(𝑣)
𝑑𝑣 = ∫
1/𝑥
𝑑𝑥

⇒ log ϕ(𝑣) = log 𝑥 + log 𝑘

⇒ log ϕ(𝑣) = log 𝑥𝑘

⇒ ϕ(𝑣) = 𝑘𝑥 ⇒ ϕ (
𝑦/𝑥
) = 𝑘𝑥 (∵ 𝑣 =
𝑥/𝑦
)

**Q10.**The solution of the differential equation 𝑑𝑦/𝑑𝑥 = sin (𝑥 + 𝑦) tan(𝑥 + 𝑦) − 1 is

Solution

Given, 𝑑𝑦/𝑑𝑥 = sin(𝑥 + 𝑦) tan(𝑥 + 𝑦) − 1

Given, 𝑑𝑦/𝑑𝑥 = sin(𝑥 + 𝑦) tan(𝑥 + 𝑦) − 1

Put 𝑥 + 𝑦 = 𝑧 ⇒ 1 +
𝑑𝑦/𝑑𝑥
=
𝑑𝑧/𝑑𝑥

∴
𝑑𝑧/𝑑𝑥
− 1 = sin 𝑧 tan 𝑧 − 1

⇒ ∫
cos 𝑧/sin²𝑧
𝑑𝑧 = ∫ 𝑑𝑥

Put sin 𝑧 = 𝑡

∴ ∫
1/𝑡² 𝑑𝑡 = 𝑥 − 𝑐

⇒ −
1/𝑡
= 𝑥 − 𝑐

⇒ −cosec 𝑧 = 𝑥 − 𝑐

⇒ 𝑥 + cosec (𝑥 + 𝑦) = c