Differential-Equation-Quiz-1

Differential Equation Quiz-1

Dear Readers,

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. The slope of a curve at any point is the reciprocal of twice the ordinate at the point and it passes

through the point (4, 3). The equation of the curve is

•  x² = 𝑦 + 5
•  y² = 𝑥 − 5
•  y² = 𝑥 + 5
•  x² = 𝑦 − 5

Solution
We have, 

Slope = 𝑑𝑦/𝑑𝑥 ⇒ 𝑑𝑦/𝑑𝑥 = 1/2𝑦 ⇒ 2 𝑦 𝑑𝑦 = 𝑑𝑥 

Integrating both sides, we get 𝑦² = 𝑥 + 𝐶 

This passes through (4, 3) 

∴ 9 = 4 + 𝐶 ⇒ 𝐶 = 5 

So, the equation of the curve is 𝑦² = 𝑥 + 5 

Q2.The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal on 𝑥-axis and passing through (2, 1) is

•  x² + y²  − 𝑥 = 0
•  4x² + 2y² − 9𝑦 = 0
•  2x² + 4y² − 9𝑥 = 0
•  4x² + 2y² − 9𝑥 = 0

Solution
∵ Equation of normal at (𝑥, 𝑦) is 

𝑌 − 𝑦 = 𝑑𝑥/𝑑𝑦 (𝑋 − 𝑥) 

Put, 𝑦 = 0 

Then, 𝑋 = 𝑥 + 𝑦 𝑑𝑦/𝑑𝑥 

Given, 𝑦² = 2𝑥 𝑋 

⇒ 𝑦² = 2𝑥 ( 𝑥 + 𝑦 𝑑𝑦/𝑑𝑥 ) 

⇒ 𝑑𝑦/𝑑𝑥 = (𝑦² − 2𝑥²)/(2𝑥𝑦) = (( 𝑦/𝑥 ) ² − 2)/ ( 2𝑦/𝑥 )

Put 𝑦 = 𝑣𝑥, we get 𝑑𝑦/𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑣/𝑑𝑥 

Then, 𝑣 + 𝑥 𝑑𝑣/𝑑𝑥 = 𝑣²−2/2𝑣 

On integrating both sides, we get 

ln (2 + 𝑣² ) + ln|𝑥| = ln 𝑐 

⇒ ln (|𝑥|(2 + 𝑣² )) = ln 𝑐 

⇒ |𝑥| ( 2 + 𝑦²/𝑥² ) = 𝑐 

∵ It passes through (2, 1), then 2 (2 + 1 4 ) = 𝑐 

⇒ 𝑐 = 9 2 

Then, |𝑥| ( 2 + 𝑦²/𝑥² ) = 9/2 

⇒ 2𝑥² + 𝑦² = 9/2 |𝑥|

⇒ 4𝑥² + 2𝑦² = 9|𝑥| 

Q3.  The differential equation obtained on eliminating 𝐴 and 𝐵 from the equation

𝑦 = 𝐴 cos 𝜔𝑡 + 𝐵 sin 𝜔𝑡 is

•  𝑦'' = −𝜔²y
•  𝑦' + 𝑦 = 0
•  𝑦'' + 𝑦' = 0
•  𝑦' − 𝜔²𝑦 = 0

Solution
Here, 𝑦 = 𝐴 cos 𝜔𝑡 + 𝐵 sin 𝜔𝑡 ….(i) 

On differentiating w. r. t.𝑡, we get 

𝑑𝑦/𝑑𝑡 = −𝜔𝐴 sin 𝜔𝑡 + 𝜔𝐵 cos 𝜔𝑡 

Again, on differentiating w. r.t.𝑡, we get 

𝑑²𝑦/𝑑𝑡² = −𝜔²𝐴cos 𝜔𝑡 − 𝜔²𝐵sin 𝜔𝑡 

⇒ 𝑑²𝑦/𝑑𝑡² = −𝜔²(𝐴 cos 𝜔𝑡 − 𝐵 sin 𝜔𝑡) 

∴ 𝑦² = −𝜔²𝑦 [from Eq. (i)]

Q4. The differential equation y(𝑑𝑦/𝑑𝑥) + x = a (a is any constant) represents

•  A set of circles having centre on the axis
•  A set of circles on the axis
•  A set of ellipses
•  None of these

Solution
We have, 

𝑦 𝑑𝑦/𝑑𝑥 + 𝑥 = 𝑎 

⇒ 𝑦 𝑑𝑦 + 𝑥 𝑑𝑥 = 𝑎 𝑑𝑥 

Integrating, we get 𝑦²/2 + 𝑥²/2 = 𝑎𝑥 + 𝐶 

⇒ 𝑥² + 𝑦² − 2 𝑎𝑥 + 2 𝐶 = 0, which represents a set of circles having centre on 𝑥-axis

Q5.The solution of 𝑑𝑦/𝑑𝑥 + 1 = cosec (𝑥 + 𝑦) is

•  cos(𝑥 + 𝑦) + 𝑥 = 𝑐
•  cos(𝑥 + 𝑦) = c
•  sin(𝑥 + 𝑦) + 𝑥 = 𝑐
•  sin(𝑥 + 𝑦) + sin(𝑥 + 𝑦) = c

Solution
 ∵ 𝑑𝑦/𝑑𝑥 + 1 = cosec (𝑥 + 𝑦) 

Let 𝑥 + 𝑦 = 𝑡 and 1 + 𝑑𝑦/𝑑𝑥 = 𝑑𝑡/𝑑𝑥 

⇒ 𝑑𝑡/(cosec 𝑡) = 𝑑𝑥 

∴ ∫ sin 𝑡 𝑑𝑡 = ∫ 𝑑𝑥 

⇒ − cos 𝑡 = 𝑥 − 𝑐 

⇒ cos(𝑥 + 𝑦) + 𝑥 = c

Q6.  If x² + y² = 1, then

•  𝑦𝑦 ′′ − (2𝑦′)² + 1 = 0
•  𝑦𝑦 ′′ + (𝑦′)² + 1 = 0
•  𝑦𝑦′′ − (𝑦′)² − 1 = 0
•  𝑦𝑦 ′′ + 2(𝑦')² + 1 = 0

Solution
Given, 𝑥² + 𝑦² = 1 

On differentiating w. r. t. 𝑥, 

we get 2𝑥 + 2𝑦𝑦 ′ = 0 

⇒ 𝑥 + 𝑦𝑦 ′ = 0 

Again , differentiating, 

we get 1 + 𝑦𝑦 ′′ + (𝑦 ′)² = 0 

Q7.The differential equation of all straight lines passing through origin is

•  𝑦 = √𝑥 𝑑𝑦/𝑑𝑥
•  𝑑𝑦/𝑑𝑥 = 𝑦 + x
•  𝑑𝑦/𝑑𝑥 = 𝑦 − 𝑥
•  None of these

Solution
The equation of all the straight lines passing through origin is 

𝑦 = 𝑚𝑥 

⇒ 𝑑𝑦/𝑑𝑥 = 𝑚 …(i) 

∴ From Eq. (i), 

𝑦 = 𝑑𝑦/𝑑𝑥 𝑥

Q8.The solution of the differential equation 𝑑𝑦/𝑑𝑥 = 𝑥 log 𝑥 is

•  𝑦 = x² log 𝑥 − x²/2 + 𝑐
•  𝑦 = (x²/2)  log 𝑥 − x²/4 + c
•  𝑦 = x²/2 + (x²/2)  log 𝑥 + 𝑐
•  None of these

Solution
Given, 𝑑𝑦 = 𝑥 log 𝑥 𝑑𝑥 

⇒ 𝑦 = 𝑥²log 𝑥/2  − ∫ 𝑥/2 𝑑𝑥 [integrating] 

⇒ 𝑦 = (x²/2)  log 𝑥 − x²/4 + c

Q9.The solution of the differential equation 𝑦𝑑𝑦/𝑑𝑥= 𝑥 − 1 satisfying 𝑦(1) = 1 is

•  y² = x² − 2𝑥 + 2
•  y² = x² − 𝑥 − 1
•  𝑦 = x² − 2𝑥 + 2
•  None of these

Solution
We have, 

𝑦 𝑑𝑦/𝑑𝑥 = 𝑥 − 1 

⇒ 𝑦 𝑑𝑦 = (𝑥 − 1)𝑑𝑥 

⇒ 𝑦²/2 = 𝑥²/2 − 𝑥 + 𝐶 

For 𝑥 = 1, we have 𝑦 = 1 

∴ 1/2 = 1/2 − 1 + 𝐶 ⇒ 𝐶 = 1 

Hence, 𝑦²/2 = 𝑥²/2 − 𝑥 + 1 ⇒ 𝑦² = 𝑥² − 2𝑥 + 2

Q10. The integral factor of equation (x² + 1) 𝑑𝑦/𝑑𝑥 + 2𝑥𝑦 = x² − 1 is 

•  x² + 1
•  2𝑥/(x² + 1)
•  (x² − 1)/(x² + 1)
•  None of these

Solution
Given, (x²+ 1) 𝑑𝑦/𝑑𝑥 + 2𝑥𝑦 = x² − 1 

or 𝑑𝑦/𝑑𝑥 + 2𝑥y/(1+x²) = (x²−1)/(x²+1) 

It is a linear differential equation. 

On comparing with the standard equation 𝑑𝑦/𝑑𝑥 + 𝑃𝑦 = 𝑄, 

we get 

𝑃 = 2𝑥/(1 +x²) ,

𝑄 = (x² − 1)/(x² + 1)