Objective Problems on Particular Integral

1. A curve that passes through (2, 4) and having subnormal of constant length of 8 units can be;

(A) y² = 16x – 8 (B) y² = -16x + 24

Solution: Let the curve be y = f(x). Subnormal at any point = | y dy/dx|

So differentiating with respect to x we get,

y dy/dx = ±8 ⇒ y dy = ±8 dx ⇒ y²/2 = ±8x + c

y² = 16 x+2c₁, if c₁= -8 or

y² = -16x +2c₂, and c₂= 24

Hence (A), (B) are correct answers.

2. Equation of a curve that would cut x² + y²- 2x - 4y - 15 = 0 orthogonally can be;

(A) (y-2) = l(x-1) (B) (y-1) = l(x-2)

Solution: Any line passing through the center of the given circle would meet the circle orthogonally.

Hence (A) is the correct answer.

3. Let m and n be the order and the degree of the differential equation whose solution is y = cx +c²- 3c^3/2 +2, where c is a parameter. Then,

(A) m = 1, n = 4 (B) m =1 , n =3

Solution: Differentiating the solution we get, dy/dx = c.

Hence, the differential equation is given by y = x dy/dx + (dy/dx)²- 3 (dy/dx)^3/2 +2.

Clearly its order is one and degree 4.

Hence (A) is the correct answer.

4. y = a sinx + b cosx is the solution of differential equation :

(A) d²y/dx² + y = 0 (B) dy/dx + y = 0

Solution: With the help of the given y, we find dy/dx and d²y/dx² and try to find the appropriate relation

dy/dx = a cos x –b sin x and d²y/dx²= -a sinx –b cosx = -y

Hence d²y/dx² +y =0.

Hence (A) is the correct answer.

5. For any differential function y = f(x), the value of d²y/dx² + (dy/dx)³ d²y/dx² is :

(A) Always zero (B) always non-zero

Solution: It is obvious that (dy/dx) = (dx/dy)^-1 for a differential equation

d²y/dx² =-1 (dx/dy)^-2 d/dy (dx/dy) dy/dx

= - (dx/dy)^-2 d²x/dy² dy/dx

= -d²x/dy²(dy/dx)³

Hence, d²y/dx² + (dy/dx)³ d²x/dy² =0.

Hence (A) is the correct answer.

6. The degree of differential equation d²y/dx² +3(dy/dx)² = x log d²y/dx² is :

(A) 1 (B) 2

Solution: Since the equation is not a polynomial in all the differential coefficients so the degree of the equation is not defined.

Hence (D) is the correct answer.

7. The degree and order of the differential equation of all parabolas whose axis is x-axis are:

(A) 2, 1 (B) 1, 2

Solution: Equation of required parabola is of the form y² = 4a(x –h)

Differentiating, we have 2y dy/dx = 4a and so, y dy/dx = 2a

Required differential equation is (dy/dx)² + d²y/dx² y =0.

Degree of the equation is 1 and order is 2.

Hence (B) is the correct answer.

8. The differential equation of all ellipses centered at origin is:

(A) y₂ + xy₁² –yy₁ = 0 (B) xyy₂ + xy₁² –yy₁ = 0

Solution: Ellipse centered at origin are given by

x²/a² + y²/b² =1 …… (1)

where a and b are unknown constants

2x/a² +2y/b² y₁ =0, where y₁ = dy/dx

x/a² +y/b² .y₁ = 0 …… (2)

Differentiating again we get, 1/a² +1/b² (y₁² +yy₂) =0 ……. (3)

Multiplying (3) with x and then subtracting from (2) we get

1/b² (yy₁ – xy₁² –xyy₂) = 0 so xyy₂ + xy₁² –yy₁ = 0.

Hence (B) is the correct answer.

9. Particular solution of y₁ + 3xy = x which passes through (0, 4) is :

(A) 3y = 1 + 11 (B) y = + 11

Solution: dy/dx + (3x) y = x

I.F =

Solution of given equation is

y + c = + c

If curve passes through (0, 4), then

4 –1/3 = c and c = 11/3

y = so 3y = 1 + 11.

Hence (A) is the correct answer.

10. Solution of equation dy/dx = (3x-4y-2)/(3x-4y-3) is :

(A) (x –y)² + c = log (3x –4y + 1) (B) x –y + c = log (3x –4y + 1)

Solution: Let 3x –4y = z

3-4dy/dx = dz/dx and dy/dx = ¼ (3-dz/dx)

Therefore the given equation ¾ - ¼ dz/dx = (z-2)/ (z-3)

Hence, (z-3) / (-z-1) dz = dx.

Solving we get, –z + 4 log (z + 1) = x + c

This gives log (3x –4y + 1) = x –y + c.

Hence (B) is the correct answer.

11. The order of the differential equation, whose general solution is

y = C₁ e^x + C₂ e^2x + C₃ e^3x + C₄ e^x+c3, where C₁, C₂, C₃, C₄, C₅ are arbitrary constants, is:

(A) 5 (B) 4

Solution: y = (c₁ + c₄) e^x + c₂ e^2x + c₃ e^3x + c₄^ec5

y = c₁e^x + c₂ e^2x + c₃ e^3x + c₄^ec5

y = k₁ e^x + k₂ e^2x + k₃ e^3x + k₄

Therefore there are 4 arbitrary constants

Hence (B) is the correct answer.

12. I.F. for y ln y dx/dy + x – ln y =0 is:

(A) ln x (B) ln y

Solution: I.F. = e ^∫^dy/dx = ln y

Hence (B) is the correct answer.

13. Which one of the following is a differential equation of the family of curves

y =Ae^2x + Be^-2x

(A) d²y/dx² -2 dy/dx +2y =0 (B) xd²y/dx² +2dy/dx –xy + x² -2 = 0

(B) d²y/dx² = 4y (D) (dy/dx)³ = 4y (x dy/dx -2y)

Solution: Given equation is y = A e^2x + b e^-2x

So dy/dx = 2 (A e^2x- b e^-2x)

d²y/dx²= 4 (Ae^2x + be^-2x) = ln y

Hence (C) is the correct answer.

14. Solution of dy/dx + sin [(x+y)/2] = sin(x-y/2) is:

(A) log tan y/4 = c – 2 sin x/2 (B) log cot y/4 = c – 2 sin x/2

Solution: dy/dx = - 2 cos x/2 sin y/2

∫ 2 cos x/2 dx = ∫ cosec y/2 dy and so c-2 sin x/2 = log tan y/4.

Hence (A) is the correct answer.

15. Solution of dy/dx = y/x + tan y/x is:

(A) sin (y/x) = kx (B) cos (y/x) = kx

Solution: dy/dx = y/x + tan y/x

put y = vx then v + x dv/dx = v + tan v

cot v dv = dx/x

Integrating, we get ln sin v = ln x + ln k

Hence, sin y/x = kx

Hence (A) is the correct answer.

16. Solution of dy/dx + √ (1-y²) / (1-x²) =0 is:

(A) sin^–1 x – sin^–1 y = c (B) sin^–1 y + sin^–1 x = c

Solution: dy/ √ (1-y²) = -dx /√ (1-x²)

So sin^-1 y + sin^-1 x = c

Hence (B) is the correct answer.

17. General solution of d²y/dx² = e^–2x is:

(A) y = ¼ e^–2x + c (B) y = e^–2x + cx + d

Solution: d²y/dx² = e^–2x

Hence, by integrating we get dy/dx = e^–2x/2+k₁

Integrating, y = e^–2x/4 + k₁x + k₂ and then y = e^–2x/4 + cx + d

Hence (C) is the correct answer.

18. Solution of dy/dx +y/x = x² is:

(A) x + y = x²/2 + c (B) x – y = x³/3 + c

Solution: dy/dx +y/x = x²

I.F. = e ^∫^dx/x = x

Therefore solution is xy = ∫x². x dx= x⁴/ 4 +c.

Hence (C) is the correct answer.

19. Differential equation associated with primitive y = Ae^3x + Be^5x:

(A) d²y/dx² -8 dy/dx +15y = 0. (B) d²y/dx² + 8 dy/dx -15y =0

Solution: y = A e^3x + B e^5x

y₁ = 3 A e^3x + 5 B e^5x

y₂= 9 A e^3x + 25 B e^5x

Therefore y₂- 8y + 15y = 0

Hence (A) is the correct answer.

20. The curve satisfying y = 2x dy/dx is a:

(A) Family of parabola (B) family of circles

Solution: dx/dy = 2dy/y

Hence, ln y = ln y² + ln c

Then y² = kx. It clearly represents a family of parabola

Hence (A) is the correct answer.

21. The equation of the curve through the origin satisfying the equation dy = (sec x + y tanx) dx is:

(A) y sin =x (B) y cosx = x

Solution: dx/dy - y tan x = sec x

I.F. = e ^{- ∫ tan x dx} = cos x

y cos x = ∫ sec x cos x dx + c

y cos x = x + c

At (0, q), y cos x = x since c = 0

Hence (B) is the correct answer.

22. The differential equation y dy/dx + x = a (where ‘a’ is a constant) represents:

(A) A set of circles having center on y-axis

(B) A set of circles with center on x-axis

(D) none of these

Solution: y dy = (a - x) dx

y²/ 2 = ax –x²/2

x² + y² = 2ax

Hence, x² + 2ax + y² = 0

It represents a set of sides with center on x-axis

Hence (B) is the correct answer.

23. If dy/dx = e^–2y and y = 0 when x = 5, then value of x where y = 3 is given by:

(A) e⁵ (B) (e⁶ +9) / 2

Solution: e^2y dy = dx

Hence, e^2y /2 + c = x

Then, c = 5 – ½ = 9/2.

At y = 3, the value of x is (e⁶ + 9) / 2

Hence (B) is the correct answer.

24. The equation of curve passing through (2,7/2) and having slope 1- 1/x² at (x, y) is:

(A) y = x² + x + 1 (B) xy = x + 1

Solution: Since the slope is given, we have dy/dx = 1- 1/x²

y = x + 1/x + c

At (2, 7/2) the value becomes 7/2 = 2 + 1/2+ c

Hence, this gives c=1.

Therefore y = x + 1/x + 1

xy = x² + x + 1

Hence (C) is the correct answer.

25. Given that dy/dx = ye^x such that x = 0, y = e. The value of y (y > 0) when x = 1 will be:

(A) e (B) 1/e

Solution: We are given that dy/dx = ye^x

Hence, dy/y = e^x dx

Integrating both sides we get, ln y = e^x + c

At x = 0, y = 1; c = 0

This gives, ln y = e^x

Therefore At x = 1, y = e^c

Hence (C) is the correct answer.

26. The differential equation of all conics with center at origin is of order

(A) 2 (B) 3

Solution: The general equation of all conics with center at origin is

ax² + 2hxy + by² = 1

Since it has three arbitrary constants, the differential equation is of order 3.

Hence (B) is the correct answer.

27. The differential equation of all conics with the coordinate axes, is of order

(A) 1 (B) 2

Solution: The general equation of all conics with axes as coordinate axes is

ax² + by² = 1.

Since it has two arbitrary constants, the differential equation is of order 2.

Hence (B) is the correct answer.

28. The differential equation of all non-vertical lines in a plane is:

(A) dy/dx = 0 (B) dx/dy =0

Solution: The general equation of non-vertical lines in a plane is

ax + by = 1, b ≠ 0

Then a+bdy/dx = 0 ⇒ dy/dx = -a/b

Integrating it again, we get d²y/dx² = 0.

Hence (C) is the correct answer.

29. The differential equation of all non-horizontal lines in a plane is:

(A) dy/dx = 0 (B) dx/dy =0

Solution: We know that the equation of non- horizontal ax = by = 0, a ≠ 0

Hence, the required differential equation will be given by d²x/dy² =0.

Hence (D) is the correct answer.

30. The degree and order of the differential equation of all tangent lines to the parabola x² = 4y is:

(A) 2, 1 (B) 2, 2

Solution: Equation of any tangent to x² = 4y is

x = my+1/m, where m is arbitrary constant …… (1)

Now differentiating this equation we get,

1= m dy/dx

⇒ m = 1/ (dy/dx)

Now putting this value of m in equation (1) we get

x = y. 1/(dy/dx) + dy/dx

Solving it we obtain the equation

(dy/dx)²–xdy/dx + y = 0.

which is clearly a differential equation of order 1 and degree 2.

Hence (A) is the correct answer.

31. If f(x) = f'(x) and f(1) = 2, then f(3) =

(A) e² (B) 2e²

Solution: Since f(x) = f'(x)

⇒ f '(x)/f(x) = 1

⇒ Integrating we get log f(x) = x + c

Since f (1) = 2

This gives log 2 = 1 + c

log f(x) = x + log 2 – 1

log f(3) = 3 + log 2 – 1

= 2 + log 2

⇒ f(3) = e^2+log2 = e² . e^log2 = 2e².

Hence (B) is the correct answer.

32. Equation of the curve passing through (3, 9) which satisfies the differential equation dy/dx = x+1/x² is:

(A) 6xy = 3x² – 6x + 29 (B) 6xy = 3x² – 29x + 6

Solution: The given differential equation is

dy/dx = x +1/x²

Integrating both sides we get, y= x²/2 -1/x + c

It passes through (3, 9)

Substituting these values in the last equation we get,

9= 9/2-1/3 +c ⇒ c=29/6.

y= x²/2 -1/x +29/6

Hence (C) is the correct answer.

33. Solution of 2y sinx dy/dx = 2sinx cosx –y²cosx, x=p/2, y=1 is given by

(A) y² = sinx (B) y = sin²x

Solution: On dividing the given equation by sin x,

2ydy/dx +y²cot x = 2cos x

Now put y² =v, then 2y dy/dx = dv/dx

Then the given equation becomes

dv/dx + v cot x = 2 cos x

I.F. = e ^{∫ cot x dx} = e^{log sinx} = sin x

Solution is v. sin x = ∫ sinx .(2 cos x) dx + c

⇒ y² sin x = sin²x + c

When x= p/2, y = 1, then c = 0

⇒ y² = sin x.

Hence (A) is the correct answer.

34. The general solution of the equation dy/dx = 1+xy is:

(A) (B)

Solution: Given equation is dy/dx = 1+xy

⇒ dy/dx –xy =1

This is the standeard form of linear equation and here, P=-x and Q=1

I.F. =

Solution is

Hence (D) is the correct answer.

35. Solution of dy/dx = e ^y+x + e ^y-x is:

(A) e^x(x + 1) = y (B) e^x(x + 1) + 1 = y

Solution: The given equation is dy/dx = e ^y+x + e ^y-x = e^y.e^x + e^y.e^-x = e^y(e^x+e^-x)

e^-y dy = (e^x +e^-x)dx

⇒ e^–ydy = (e^x + e^–x)dx

Integrating we get, e^–y= e^x – e^–x + c

Hence (D) is the correct answer.

36. If y = e^–x (A cosx + B sin x), then y satisfies

(A) d²y/dx² + 2dy/dx =0 =0 (B) d²y/dx² - 2dy/dx +2y =0

Solution: y = e^–x (A cos x + B sin x) ....(1)

⇒ dy/dx = e^-x (-A sin x +B cos x) - e^-x (A cos x +B sin x)

⇒ dy/dx = e^-x (-A sinx +B cos x) –y ……(2)

⇒ d²y/dx² = e^-x(-A cosx -B sin x) –e^-x (-A sinx +B cos x)

Using (1) and (2), we get

d²y/dx² +dy/dx = -y-y-dy/dx

d²y/dx² +2dy/dx + 2y= 0.

Hence (C) is the correct answer.

37. The solution of x² dy/dx –xy = 1+ cos y/x is:

(A) tan y/2x = c-1/2x² (B) tan y/x = c+1/x

Solution: Dividing the given equation by x² throughout

dy/dx –y/x = 1/x²+ 1/x² cos y/x .....(1)

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

Then equation (1) becomes v + x dv/dx –v =1/x² +1/x²cos v

⇒ x³ dv/dx = 1+cos v

Then dv/ (1+cos v) = dx/x³

Integrating both sides, ∫dv/ (1+cos v) = ∫dx/x³

½ sec²v/2 dv = -1/x +c

⇒ tan v/2 = -1/2x² +c

Hence (A) is the correct answer.

38. The order of differential equation d²y/dx² = 1+dy/dx is:

(A) 2 (B 3

Solution: Clearly the equation is of order 2.

Hence (A) is the correct answer.

39. The solution of the differential equation 2x dy/dx –y =3 represents

(A) straight lines (B) circles

Solution: The given equation is 2x dy/dx =y+3.

Hence, Separating the terms of x and y, we get

⇒ 2dy/(y+3) = dx/x

Integrating, 2log (y+3) = log x +log c

⇒ (y + 3)² = ex, which represents a parabola.

Hence (C) is the correct answer.

40. Solution of differential equation dy – sinx siny dx = 0 is:

(A) e^cosx tan y/2 = c (B) e^cosx tan y = c

Solution: Given equation can be written as:

dy/ siny = sinx dx

Now integrating both sides we get,

∫ cosec y dy = ∫ sinx dx +c

log tan y/2 = -cos x+c₁

tan y/2 = e^{-cos x} e^c₁

⇒ e ^cosx tan y/2 = c

Hence (A) is the correct answer.

41. Solution of differential equation dy/dx + ay =e^mx is:

(A) (a + m) y = e^mx + c (B) ye^ax = me^mx + c

Solution: The given equation is dy/dx + ay =e^mx

This linear equation is already in the standard form and so P= a, Q= e^mx

I.F. = e ^{∫ adx} = e^ax

Hence, the solution is y.e^ax = ∫ e^ax. e^mx dx +c

⇒ y e^ax =e^(a+m)x/ (a+m) +c₁

⇒ (a + m) y = e^mx + c₁ (a + m) e^–ax

⇒ (a + m) y = e^mx + ce^–ax, where c = c₁ (a + m)

Hence (D) is the correct answer.

42. Integrating factor of the differential equation cos x dy/dx +y sin x=1 is:

(A) cos x (B) tan x

Solution: The given equation is cos x dy/dx + y sin x=1

This can be rewritten as dy/dx + y sin x /cos x = 1/cos x

Here P= tan x, Q= sec x

I.F. = e ^{∫ tan x dx} = e^{log secx} = sec x.

Hence (C) is the correct answer.

43. Solution of differential equation xdy – ydx = 0 represents

(A) A rectangular hyperbola

(B) Straight line passing through origin

(D) Circle whose center is at origin

Solution: Given equation is x dy – y dx = 0

⇒ dy/y –dx/x =0

⇒ log y – log x = log c

⇒ log y/x = log c

⇒ y/x =c or y=cx which is the equation of a straight line.

Hence (B) is the correct answer.

Watch this Video for more reference

44. The integrating factor of the differential equation

dy/dx (x log x) + y =2log x is given by

(A) e^x (B) log x

Solution: The given equation is

dy/dx (x log x) + y =2log x

Dividing throughout by xlog x the equation reduces in the form,

dy/dx + y/xlog x = 2/x

⇒ P =1/xlog x, Q = 2/x

Hence (B) is the correct answer.

45. The solution of dy/dx +1 = e ^x+y is:

(A) (x + y) e^x+y = 0 (B) (x + c) e^x+y = 0

Solution: The given equation is dy/dx + 1= e^x+y

⇒ dy/dx +1 = e^x.e^y

⇒ e^-y dy/dx +e^-y = e^x

Put e^–y = z

⇒ -e^–ydy/dx = dz/dx

Hence, the given equation becomes dz/dx - z = -e^x

This is clearly a linear equation where P= -1 and Q= -e^x

I.F. = e ^∫–1dx = e^–x

Solution is z e^–x = ∫ –e^x . e^–x dx + c = –x + c

⇒ e^–y .e^–x = c – x

⇒ e^–(x+y) = c – x

⇒ (x – c) e^x+y + 1 = 0

Hence (D) is the correct answer.

46. Family y = Ax + A³ of curves is represented by the differential equation of degree

(A) 3 (B) 2

Solution: There is only one arbitrary constant iinvolved and hence the degree of the differential equation is one.

Hence (C) is the correct answer.

47. Integrating factor of dy/dx +y/x = x³; (x > 0) is :

(A) x (B) log x

Solution: I.F. is given by exp( ∫1/x dx) = x.

Hence (A) is the correct answer.

48. If integrating factor of x(1 – x²) dy + (2x²y – y – ax³) dx = 0 is e ^{∫ Pdx}, then P is:

(A) (2x² –ax³) / (x (1-x²)) (B) 2x³ – 1

Solution: The given equation is x(1 – x²) dy + (2x²y – y – ax³) dx = 0.

Hence ⇒ x(1 – x²) dy/dx + 2x²y – y – ax³ = 0.

⇒ dy/dx + {(2x² -1)/ x (1-x²)} . y = ax³/ x (1-x²)

Hence, clearly P is (2x³-1)/ x (1-x²).

Hence (D) is the correct answer.

49. For solving dy/dx =4x +y+1, suitable substitution is:

(A) y = vx (B) y = 4x + v

Solution: The given equation can be solved by substituting y + 4x + 1 = v

Hence (D) is the correct answer.

50. Integral curve satisfying y'= (x²+y²)/ (x²-y²), y (1) =1 has the slope at the point (1, 0) of the curve equal to:

(A) -5/3 (B) –1

Solution: dy/dx = (x²+y²)/ (x²-y²)

Slope at (1,0) = (dy/dx)_(1,0) = 1+0/1-0 =1.

Hence (C) is the correct answer.

51. A solution of the differential equation is :

(A) y = 2 (B) y = 2x

Solutions : (C)

52. The solution of (2x-10y³) dx/dy + y=0 is:

(A) x + y = ce^2x (B) y² = 2x³ + c

Solution: We can rewrite the given equation in the form

ydy/dx= -2x +10y³

⇒ dx/dy +2x/y = 10y²

This is a linear equation with P= 2/y and Q= 10y²

Hence, I.F. exp (2/y dy) = e ^{2log y} = exp (log y²) = y²

Hence, the solution is given by xy² = ∫ 10 y².y² dy +c

⇒ xy²= 10y⁵/5 +c.

Hence (C) is the correct answer.

53. A particle moves in a straight line with velocity given by dx/dt =x+1, where x denotes the distance described. The time taken by the particle to describe 99 metres is:

(A) log₁₀e (B) 2 log_e 10

Solution: The given equation is dx/dt =x+1. It may be written as

dx/(x+1) = dt

⇒ log(x + 1) = t + c ....(1)

Initially, when t = 0, x = 0

⇒ c=0.

⇒ log (x + 1) = t

When x = 99, then t = log_e (100) = 2 log_e 10.

Hence (B) is the correct answer.

54. The curve for which slope of the tangent at any point equals the ratio of the abscissa to the ordinate of the point is a/an

(A) Ellipse (B) rectangular hyperbola

Solution: The equation of the required curve is dy/dx = x/y

⇒ ydy = xdx

Integrating both sides, y²/2 –x²/2 =c

⇒ y² – x² = 2c

which is a rectangular hyperbola

Hence (B) is the correct answer.

55. The solution of the differential equation (1 – xy – x⁵y⁵) dx – x²(x⁴y⁴ + 1)dy = 0 is given by

(A) x = (B) x =

Solution: The given equation is dx – x (ydx+ xdy) = x⁵y⁴(ydx + xdy)

⇒ dx/x = (1+x⁴y⁴) d (xy)

⇒ lnx = xy + 1/5 x⁵y⁵ + ln c

⇒ x = c exp (xy+x⁵y⁵/5)

Hence (A) is the correct answer.

56. If y= e^4x +2e^-x satisfies the relation d³y/dx³ + A dy/dx + By= 0 then value of A and B respectively are:

(A) –13, 14 (B) –13, –12

Solution: On differentiating y= e^4x +2e^-x, w.r.t x we get

dy/dx = 4e^4x – 2e^-x

⇒ d²y/dx² = 16e^4x + 2e^-x

⇒ d³y/dx³ = 64e^4x– 2e^-x

Putting these values in d³y/dx³ + A dy/dx + By= 0 we get,

(64+4A+B)e^4x +(-2-2A+2B)e^-x =0.

On solving we get A= –13 and B = –12

Hence (B) is the correct answer.

57. The differential equation of all circles which pass through the origin and whose centers lie on y-axis is:

(A) (x²-y²)dy/dx -2xy =0 (B) (x²-y²)dy/dx+2xy =0

Solution: If (0, a) is the center on y-axis, then its radius is ‘a’ because it passes through origin.

Equation of circle is x² + (y – a)² = a²

⇒ x² + y² – 2ay = 0 ....(1)

⇒ 2x+2ydy/dx -2ady/dx =0. .....(2)

Using (1) in (2), 2x+2ydy/dx – (x²+y²)/y .dy/dx =0.

⇒ 2xy = (x²+y²-2y²)dy/dx

⇒ dy/dx = 2xy/ (x²-y²)

Hence (A) is the correct answer

58. If xdy/dx = y (logy – logx +1) then the solution of the equation is:

(A) log x/ y = cy (B) log y/ x = cy

Solution: The given equation is xdy/dx = y (logy – logx +1)

⇒ dy/dx= y/x (log y/x +1)

Put y=vx

Then dy/dx= v+xdv/dx

The equation reduces to v+xdv/dx = v(log v+1)

⇒ dv/ v log v = dx/x

⇒ log(log v) = logx + logc = logcx

⇒ log v = cx.

Hence (D) is the correct answer.

The methods of finding the particular integral are quite easy and the topic fetches 2-3 questions in the JEE exam. You may also refer the Sample Papers to get an idea about the types of questions asked in the exam.

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