# NCERT MCQ’s : Class 12 Maths with Answers MCQ Chapter 9 Differential Equations

## NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations MCQ’s with Answers

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## NCERT Maths MCQs for Class 12 Question and Answers Chapter  9 Differential Equations PDF

(a) 1
(b)2
(c) 3
(d) 4

Option b – 2

### Q2. The differential equation of the family of lines passing through ongrn is

(a) y = mx
(b) dy/dx = m
(c) x dy – y dx = 0
(d) dy/dx = 0

Option c – x dy – y dx = 0

### Q3. Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of the x-axis.

Option – General equation of parabola is y² = 4ax …..(i) Differentiating, we get 2yy’ = 4a ⇒ yy’ = 2a. Substituting in (i), we gety2 = 2xyy’.

### Q4. Find the particular solution of the differential equation dy/dx =y tanx, given that y= 1 when x = 0.

Option – ∫ dy/dx = ∫tan x dx ⇒ log |y| = log|sec x| + log C ⇒ y = C sec x ….(i) Given y = 1, x = 0 ⇒ 1 = C sec 0 ⇒ C = 1 ∴ solution is y = sec x [from (i)]

Q5.

Option c –

(a) cos x
(b) sec x
(c) ecos x
(d) esec x

Option b – sec x

### Q7. If (x + y)2 dy/dx = a2, y = 0 when x = 0, then y = a if x/a =

(a) 1
(b) tan 1
(c) tan 1 + 1
(d) tan 1 – 1

Option d – (d) tan 1 – 1

### Q8. If sinx dy/dx + y cosx = x sinx, then (y – 1) sinx =

(a) c – x sinx
(b) c + xcosx
(c) c – x cos x
(d) c + x sin x

Option c – c – x cos x

### Q9. The solution of differential equation (ey + 1) cosx dx + ey sinx dy = 0 is

(a) (ey + 1) sinx = c
(b) ex sinx = c
(c) (ex + 1) cosx = c
(d) none of these

Option a – (ey + 1) sinx = c

Option – a

### Q11. The differential equation representing the family of ellipses with centre at origin and foci on x-axis is given as

(a) xy’ + y = 0
(b) x2y2(y”)2 + yy’= 0
(c) xyy” + x(y’)2 – yy’ = 0
(d) None of these

Option b – x2y2(y”)2 + yy’= 0

### Q12. The differential equation satisfied by y = A/x + B is (A, B are parameters)

(a) x2 y1 = y
(b) xy1 + 2y2 = 0
(c) xy2 + 2y1 = 0
(d) none of these

Option c – xy2 + 2y1 = 0

### Q13. The particular solution In(dy/dx) = 3x + 4y, y(0) = 0 is

(a) e3x + 3e-4y = 4
(b) 4e3x – 3e-4y = 3
(c) 3e3x + 4e4y = 7
(d) 4e3x + 3e-4y = 7

Option d – 4e3x + 3e-4y = 7

### Q14. The solution of differential equation dy/dx =x-y/x+y is

(a) x2 – y2 + 2xy + c = 0
(b) x2 – y2 – xy + c = 0
(c) x2 – y2 + xy + c = 0
(d) x2 – y2 – 2xy + c = 0

Option d – x2 – y2 – 2xy + c = 0

(a) 3
(b) 2
(c) 1
(d) 5

Option a – 3

### Q16. The differential equation of all parabolas whose axis of symmetry is along the axis of the x-axis is of order

(a) 3
(b) 1
(c) 2
(d) none of these

Option c – 2

### Q17. The solution of the differential equation cos x sin y dx + sin x cos y dy = 0 is

(a) sinx/siny=c
(b) sin x sin y = c
(c) sin x + sin y = c
(d) cos x cos y = c

Option b – sin x sin y = c

(a) 3
(b) 2
(c) 1
(d) 5

Option a – 3

### Q19. The radius of a circle is increasing at the rate of 0.4 cm/ s. The rate of increasing of its circumference is

(a) 0.4 π cm/s
(b) 0.8 π cm/s
(c) 0.8 cm/s
(d) None of these

Option b – 0.8 π cm/s

### Q20. Solution of differential equation xdy – ydx = Q represents

(a) a rectangular hyperbola
(b) parabola whose vertex is at origin
(c) straight line passing through origin
(d) a circle whose centre is at origin

Option c – straight line passing through origin

(a) 1
(b) 2
(c) 3
(d) 4

Option b – 2

### Q22. The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is

(a) an ellipse
(b) parabola
(c) circle
(d) rectangular hyperbola

Option d – rectangular hyperbola

(a) 3
(b) 2
(c) 1
(d) 0

Option d – 0

### Q24. Which of the following is a homogeneous differential equation?

(a) (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0
(b) xy dx – (x³ + y²)dy = Q
(c) (x³ + 2y²) dx + 2xy dy = 0
(d) y² dx + (x² – xy – y²)dy = 0

Option d – y² dx + (x² – xy – y²)dy = 0

### Q25. The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is

(a) (x – 2)² y’² = 25 – (y – 2)²
(b) (x – 2) y’² = 25 – (y – 2)²
(c) (y – 2) y’² =25 – (y – 2)²
(d) (y – 2)² y’² = 25 – (y – 2)²

Option d – (y – 2)² y’² = 25 – (y – 2)²