Mathematics MCQ on Differential Equations for JEE and Engineering Exam 2022

MCQ on Differential Equations: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.

MCQ on Differential Equations

In this post we are providing you MCQ on Differential Equations, which will be beneficial for you in upcoming JEE and Engineering Exams.

MCQ on Differential Equations

Q1. Order of the differential equation of the family of all concentric circles centred at (h, k) is _______.
a) -1
b) 1
c) 0
d) None of these

View Answer

Solution:

(x − h)2 + (y − k)= r2

Here r is arbitrary constant.

Order of differential equation = 1.

Q2. (x+ y2) dy = xy dx. If y (x0) = e, y (1) = 1, then the value of x0 = __________.
a)  √3e
b) ± √3e
c) ± √2e
d) none

View Answer

Solution:

(x+ y2) dy = xy dx

x (x dy − y dx) = −y2 dy

[x * (y dx − x dy)] / y2 = dy

[x / y] d (x / y) = dy / y

Integrating, x/ 2y2 = loge y + c

Given y (1) = 1

c = 1 / 2

x/ 2y= loge y + 1 / 2

Now y (x0) = e

x0/ 2e− loge e − 1 / 2 = 0

x0= 3e2

x= ± √3e

Q3. Solution of differential equation 2xy dy / dx = x2 + 3y2 is __________.
a) x2 + y= px
b) x2 + y= px3
c) x2 + y= px2
d) None of these

View Answer

Solution:

It is homogeneous equation dy / dx = [x2 + 3y2] / 2xy

Put y = vx and dy / dx = v + x * [dv / dx]

So, we get x * [dv / dx] = [1 + v2] / 2v

2v dv / 1 + v2 = dx / x

On integrating, we get

x2 + y= px3. (where p is a constant)

Q4.  The equation of the curve which passes through the point (1, 1) and whose slope is given by 2y / x, is __________.
a) y = x3
b) y = x2
c) y = x
d) None of these

View Answer

Solution:

Slope dy / dx = 2y / x

2 ∫dx / x =∫dy / y

2 log x = log y + log c

x= yc

Since it passes through (1, 1), therefore, c = 1.

Hence, x− y = 0

y = x2.

Q5. The general solution of the differential equation (x + y) dx + x dy = 0 is _______.
a) y+ 2xy = c.
b) x2  -2xy = c.
c) x+ 2xy = c.
d) None of these

View Answer

Solution:

(x + y) dx + x dy = 0

x dy = − (x + y) dx

dy / dx = −[x + y] / x

It is homogeneous equation, hence, put y = vx and dy / dx = v + x [dv / dx], we get

v + x dv / dx = −[x + vx] / x= −[1 + v] / 1

x dv / dx = −1 −2v

∫dv / [1 + 2v] = −∫dx / x

[1 / 2] log (1 + 2v) = −log x + log c

log (1 + 2 [y / x]) = 2 log [c / x] [x + 2y] / x = (c / x)2

x+ 2xy = c.

Q6. A function y = f (x) has the second-order derivatives f′′(x) = 6 (x − 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x − 5, then the function is __________.
a) (x + 1)3.
b) (x − 1)2.
c) (x − 1)3.
d) (x + 1)2.

View Answer

Solution:

Given f′′(x) = 6 (x − 1)

f′(x) = 3 (x − 1)+ c1 ..(i)

But at point (2, 1) the line y = 3x − 5 is tangent to the graph y = f(x).

Hence, dy / dx|x = 2 = 3 or

f′ (2) = 3.

Then from (i) f′ (2) = 3 (2 − 1)+ c1

3 = 3 + c1

c= 0 i.e.,

f′ (x) = 3 (x − 1)2

Given f (2) = 1

f (x) = (x − 1)+ c2

f (2) = 1 + c2

1 = 1 + c2

c= 0

Hence, f (x) = (x − 1)3.

Q7. The differential equation of all parabolas whose axes are parallel to y-axis is ________________.
a) d3y / dx3 = 1
b) d3y / dx3 = 0
c) d3y / dx3 = -1
d) None of these

View Answer

Solution:

The equation of a member of the family of parabolas having an axis parallel to y-axis is

y = Ax+ Bx + C …..(i) where A, B, C are arbitrary constants.

Differentiating (i) w.r.t. x, we get

dy / dx = 2Ax + B …..(ii) which on differentiating w.r.t. x gives

dy / dx2 = 2A …..(iii)

Differentiating w.r.t. x again, we get

d3y / dx3 = 0.

Q8. Equation of curve through the point (1, 0) which satisfies the differential equation (1 + y2) dx − xy dy = 0, is ________.

View Answer

We have [dx / x] = [y dy] / [1 + y2]

Integrating, we get log |x| = ([1 / 2] * log [1 + y2]) + log c

|x| = c √(1 + y2)−−−−−−−

But it passes through (1, 0), so we get c = 1

Therefore, the solution is x2 = y2 + 1 or x2 − y= 1

Q9. The differential equation corresponding to primitive y = ecx is, or the elimination of the arbitrary constant m from the equation y = emx gives the differential equation ________.
a)  (X / Y) log X
b)  (X / Y) log y
c) (y / x) log y
d)  (y / x) log X

View Answer

Solution:

y = emx

logy = mx ⇒ m = [log y] / x

Now y = emx

dy / dx = memx

= ([log y] / x) * y

= (y / x) log y

Q10. If x dy = y (dx + y dy), y > 0 and y (1) = 1, then y (−3) is equal to _________.
a) -3
b) 0
c) 3
d) None of these

View Answer

Solution:

x dy = y (dx + y dy)

[x dy − ydx] / [y2] = dy

−d (x / y) = dy

Integrating both sides, we get x / y + y = c [Because y (1) = 1 ⇒ c = 2; Hence xy + y = 2 ]

For x = −3,

y− 2y − 3 = 0

⇒ y = −1 or 3

⇒y = 3 (Because y > 0)


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