## Differential Equation Quiz-2

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
•  x
•  𝑥²⁄2
•  1⁄x
•  1/𝑥²
Solution
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
•  (𝑥² − 𝑦² ) 𝑑𝑦/𝑑𝑥 − 2𝑥𝑦 = 0
•  (𝑥² − 𝑦²  ) 𝑑𝑦/𝑑𝑥 + 2𝑥𝑦 = 0
•  (𝑥² − 𝑦²  ) 𝑑𝑦/𝑑𝑥 − 𝑥𝑦 = 0
•  (𝑥² − 𝑦²  ) 𝑑𝑦/𝑑𝑥 + 𝑥𝑦 = 0
Solution
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 = 𝑎ebx  is
•  𝑦 𝑑 2𝑦 𝑑𝑥2 + 𝑑𝑦 𝑑𝑥 = 0
•  𝑦 𝑑 2𝑦 𝑑𝑥2 − 𝑑𝑦 𝑑𝑥 = 0
•  𝑦 𝑑 2𝑦 𝑑𝑥 2 − ( 𝑑𝑦 𝑑𝑥) 2 = 0
•  𝑦 𝑑 2𝑦 𝑑𝑥 2 + ( 𝑑𝑦 𝑑𝑥) 2 = 0
Solution
The given equation is 𝑦 = 𝑎ebx
⇒ 𝑑𝑦/𝑑𝑥 = 𝑎𝑏ebx  …(i)
⇒ 𝑑²𝑦/𝑑𝑥² = 𝑎𝑏 2ebx  …(ii)
⇒ 𝑎ebx 𝑑 2𝑦 𝑑𝑥2 = 𝑎²𝑏² ebx ebx
⇒ 𝑦 𝑑²𝑦/𝑑𝑥² = ( 𝑑𝑦/𝑑𝑥)² [from eq. (ii)]

Q4. A continuously differential function ϕ(𝑥) in (0, 𝜋) satisfying 𝑦 ′ = 1 + 𝑦
² ,𝑦(0) = 0 = 𝑦(𝜋), is
•  tan 𝑥
•  𝑥(𝑥 − 𝜋)
•  (𝑥 − 𝜋)(1 − 𝑒 𝑥 )
•  Not possible
Solution
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
•  A parabola whose vertex is at the origin
•  A circle whose centre is at the origin
•  A rectangular hyperbola
•  Straight lines passing through the origin
Solution
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
•  First order and second degree
•  First order and first degree
•  Second order and first degree
•  Second order and second degree
Solution
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
•  1
•  3
•  4
•  None of these
Solution
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
•  ϕ ( 𝑦/𝑥 ) = 𝑘x
•  𝑥ϕ ( 𝑦/𝑥 ) = 𝑘
•  ϕ ( 𝑦/𝑥 ) = 𝑘𝑦
•   𝑦ϕ ( 𝑦/𝑥 ) = 𝑘
Solution
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
•  cosec (𝑥 + 𝑦) + tan(𝑥 + 𝑦) = 𝑥 + c
•  𝑥 + cosec (𝑥 + 𝑦) = c
•  𝑥 + tan(𝑥 + 𝑦) = 𝑐
•  𝑥 + sec(𝑥 + 𝑦) = 𝑐
Solution
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 #### Written by: AUTHORNAME

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