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.
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
Solution:
(x − h)^{2} + (y − k)^{2 }= r^{2}
Here r is arbitrary constant.
Order of differential equation = 1.
Q2. (x^{2 }+ y^{2}) dy = xy dx. If y (x_{0}) = e, y (1) = 1, then the value of x_{0} = __________.
a) √3e
b) ± √3e
c) ± √2e
d) none
Solution:
(x^{2 }+ y^{2}) dy = xy dx
x (x dy − y dx) = −y^{2} dy
[x * (y dx − x dy)] / y^{2} = dy
[x / y] d (x / y) = dy / y
Integrating, x^{2 }/ 2y^{2} = loge y + c
Given y (1) = 1
c = 1 / 2
x^{2 }/ 2y^{2 }= loge y + 1 / 2
Now y (x_{0}) = e
x_{0}^{2 }/ 2e^{2 }− loge e − 1 / 2 = 0
x_{0}^{2 }= 3e^{2}
x_{0 }= ± √3e
Q3. Solution of differential equation 2xy dy / dx = x^{2} + 3y^{2} is __________.
a) x^{2} + y^{2 }= px
b) x^{2} + y^{2 }= px^{3}
c) x^{2} + y^{2 }= px^{2}
d) None of these
Solution:
It is homogeneous equation dy / dx = [x^{2} + 3y^{2}] / 2xy
Put y = vx and dy / dx = v + x * [dv / dx]
So, we get x * [dv / dx] = [1 + v^{2}] / 2v
2v dv / 1 + v^{2} = dx / x
On integrating, we get
x^{2} + y^{2 }= px^{3}. (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 = x^{3}
b) y = x^{2}
c) y = x
d) None of these
Solution:
Slope dy / dx = 2y / x
2 ∫dx / x =∫dy / y
2 log x = log y + log c
x^{2 }= yc
Since it passes through (1, 1), therefore, c = 1.
Hence, x^{2 }− y = 0
y = x^{2}.
Q5. The general solution of the differential equation (x + y) dx + x dy = 0 is _______.
a) y^{2 }+ 2xy = c.
b) x^{2 } -2xy = c.
c) x^{2 }+ 2xy = c.
d) None of these
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^{2 }+ 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}.
Solution:
Given f′′(x) = 6 (x − 1)
f′(x) = 3 (x − 1)^{2 }+ c_{1} ..(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)^{2 }+ c_{1}
3 = 3 + c_{1}
c_{1 }= 0 i.e.,
f′ (x) = 3 (x − 1)^{2}
Given f (2) = 1
f (x) = (x − 1)^{3 }+ c_{2}
f (2) = 1 + c_{2}
1 = 1 + c_{2}
c_{2 }= 0
Hence, f (x) = (x − 1)^{3}.
Q7. The differential equation of all parabolas whose axes are parallel to y-axis is ________________.
a) d^{3}y / dx^{3} = 1
b) d^{3}y / dx^{3} = 0
c) d^{3}y / dx^{3} = -1
d) None of these
Solution:
The equation of a member of the family of parabolas having an axis parallel to y-axis is
y = Ax^{2 }+ 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
d^{2 }y / dx^{2} = 2A …..(iii)
Differentiating w.r.t. x again, we get
d^{3}y / dx^{3} = 0.
Q8. Equation of curve through the point (1, 0) which satisfies the differential equation (1 + y^{2}) dx − xy dy = 0, is ________.
We have [dx / x] = [y dy] / [1 + y^{2}]
Integrating, we get log |x| = ([1 / 2] * log [1 + y^{2}]) + log c
|x| = c √(1 + y^{2})−−−−−−−
But it passes through (1, 0), so we get c = 1
Therefore, the solution is x^{2} = y^{2} + 1 or x^{2} − y^{2 }= 1
Q9. The differential equation corresponding to primitive y = e^{cx} is, or the elimination of the arbitrary constant m from the equation y = e^{mx }gives the differential equation ________.
a) (X / Y) log X
b) (X / Y) log y
c) (y / x) log y
d) (y / x) log X
Solution:
y = e^{mx}
logy = mx ⇒ m = [log y] / x
Now y = e^{mx}
dy / dx = me^{mx}
= ([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
Solution:
x dy = y (dx + y dy)
[x dy − ydx] / [y^{2}] = 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^{2 }− 2y − 3 = 0
⇒ y = −1 or 3
⇒y = 3 (Because y > 0)
MCQ Link for NEET/JEE | |
JEE/NEET Physics MCQ | Click Here |
NEET/JEE Chemistry MCQ | Click Here |
NEET Biology MCQ | Click Here |
JEE Math’s MCQ | Click Here |
Notes PDF Link for NEET/JEE | |
Physics Notes PDF | Click Here |
Chemistry Notes PDF | Click Here |
Biology Notes PDF | Click Here |
Math’s Notes PDF | Click Here |
Follow on Facebook
By Team Learning Mantras