**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) y^{2} = 16x – 8 (B) y^{2} = -16x + 24

(C) x^{2} = 16y – 60 (D) x^{2} = -16y + 68

**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 ⇒ y^{ 2}/2 = ±8x + c

y^{2} = 16 x+2c_{1}, if c_{1}= -8 or

y^{2} = -16x +2c_{2}, and c_{2}= 24

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

**2. Equation of a curve that would cut x ^{2} + y^{2 }- 2x - 4y - 15 = 0 orthogonally can be**;

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

(C) (y+2) = l(x+1) (D) (y+1) = l(x+2) where l ∈R.

**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 +c ^{2}- 3c^{3/2} +2, where c is a parameter. Then,**

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

(C) m = 1, n = 2 (D) None

**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) d^{2}y/dx^{2} + y = 0 (B) dy/dx + y = 0

(C) d^{2}y/dx^{2} = y (D) dy/dx = y

**Solution:** With the help of the given y, we find dy/dx and d^{2}y/dx^{2} and try to find the appropriate relation

dy/dx = a cos x –b sin x and d^{2}y/dx^{2 }= -a sinx –b cosx = -y

Hence d^{2}y/dx^{2} +y =0.

Hence (A) is the correct answer.

**5. For any differential function y = f(x), the value of d ^{2}y/dx^{2} + (dy/dx)^{3} d^{2}y/dx^{2} is :**

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

(C) equal to 2y^{2} (D) equal to x^{2}

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

d^{2}y/dx^{2} =-1 (dx/dy)^{-2} d/dy (dx/dy) dy/dx

= - (dx/dy)^{-2} d^{2}x/dy^{2} dy/dx

= -d^{2}x/dy^{2 }(dy/dx)^{3}

Hence, d^{2}y/dx^{2} + (dy/dx)^{3} d^{2}x/dy^{2} =0.

Hence (A) is the correct answer.

**6. The degree of differential equation d ^{2}y/dx^{2} +3(dy/dx)^{2} = x log d^{2}y/dx^{2} is :**

(A) 1 (B) 2

(C) 3 (D) none of these

**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

(C) 3, 2 (D) none of these

**Solution:** Equation of required parabola is of the form y^{2} = 4a(x –h)

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

Required differential equation is (dy/dx)^{2} + d^{2}y/dx^{2} 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) y_{2} + xy_{1}^{2} –yy_{1} = 0 (B) xyy_{2} + xy_{1}^{2} –yy_{1} = 0

(C) yy_{2} + xy_{1}^{2} –xy_{1} = 0 (D) none of these

**Solution:** Ellipse centered at origin are given by

x^{2}/a^{2} + y^{2}/b^{2} =1 …… (1)

where a and b are unknown constants

2x/a^{2} +2y/b^{2} y_{1} =0, where y_{1} = dy/dx

x/a^{2} +y/b^{2} .y_{1} = 0 …… (2)

Differentiating again we get, 1/a^{2} +1/b^{2} (y_{1}^{2} +yy_{2}) =0 ……. (3)

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

1/b^{2} (yy_{1} – xy_{1}^{2} –xyy_{2}) = 0 so xyy_{2} + xy_{1}^{2} –yy_{1} = 0.

Hence (B) is the correct answer.

**9. Particular solution of y _{1} + 3xy = x which passes through (0, 4) is :**

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

(C) y = 1 + (D) 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)

(C) x –y + c = = log (3x –4y –3) (D) 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 = C_{1} e^{x} + C_{2} e^{2x} + C_{3} e^{3x} + C_{4} e^{x+c3}, where C_{1}, C_{2}, C_{3}, C_{4}, C_{5} are arbitrary constants, is:

(A) 5 (B) 4

(C) 3 (D) none of these

**Solution:** y = (c_{1} + c_{4}) e^{x} + c_{2} e^{2x} + c_{3} e^{3x} + c_{4}^{ec5}

y = c_{1 }e^{x} + c_{2} e^{2x} + c_{3} e^{3x} + c_{4}^{ec5}

y = k_{1} e^{x} + k_{2} e^{2x} + k_{3} e^{3x} + k_{4}

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

(C) ln xy (D) none of these

**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) d^{2}y/dx^{2} -2 dy/dx +2y =0 (B) xd^{2}y/dx^{2} +2dy/dx –xy + x^{2} -2 = 0

(B) d^{2}y/dx^{2} = 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})

d^{2}y/dx^{2 }= 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

(C) log tan y/4 = c – 2 cos x/2 (D) none of these

**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

(C) tan (y/x) = kx (D) none of these

**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-y ^{2}) / (1-x^{2}) =0 is:**

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

(C) sin^{–1} x = c sin^{–1} y (D) (sin^{–1} x) (sin^{–1} y) = c

Solution: dy/ √ (1-y^{2}) = -dx /√ (1-x^{2})

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

Hence (B) is the correct answer.

**17. General solution of d ^{2}y/dx^{2} = e^{–2x} is:**

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

(C) y = ¼ e^{–2x} + cx + d (D) y = e^{–2x} + cx^{2} + d

**Solution:** d^{2}y/dx^{2} = e^{–2x}

Hence, by integrating we get dy/dx = e^{–2x}/2+k_{1}

Integrating, y = e^{–2x}/4 + k_{1}x + k_{2 } and then y = e^{–2x}/4 + cx + d

Hence (C) is the correct answer.** **

**18. Solution of dy/dx +y/x = x ^{2} is:**

(A) x + y = x^{2}/2 + c (B) x – y = x^{3}/3 + c

(C) xy = x^{4} + c (D) y – x = x^{4} + c

**Solution:** dy/dx +y/x = x^{2}

I.F. = e ^{∫}^{ dx/x} = x

Therefore solution is xy = ∫x^{2}. x dx= x^{4 }/ 4 +c.

Hence (C) is the correct answer.

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

(A) d^{2}y/dx^{2} -8 dy/dx +15y = 0. (B) d^{2}y/dx^{2} + 8 dy/dx -15y =0

(C) d^{2}y/dx^{2} + 8 dy/dx + y =0 (D) none of these

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

y_{1} = 3 A e^{3x} + 5 B e^{5x}

y_{2}= 9 A e^{3x} + 25 B e^{5x}

Therefore y_{2}- 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

(C) Pair of straight line (D) none of these

**Solution:** dx/dy = 2dy/y

Hence, ln y = ln y^{2} + ln c

Then y^{2} = 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

(C) y tanx = x (D) none of these

**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

(C) A set of ellipses

(D) none of these

**Solution:** y dy = (a - x) dx

y^{2}/ 2 = ax –x^{2}/2

x^{2} + y^{2} = 2ax

Hence, x^{2} + 2ax + y^{2} = 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) e^{5} (B) (e^{6} +9) / 2

(C) e^{6} + 1 (D) log_{e} 6

**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 (e^{6} + 9) / 2

Hence (B) is the correct answer.** **

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

(A) y = x^{2} + x + 1 (B) xy = x + 1

(C) xy = x^{2} + x + 1 (D) xy = y + 1

**Solution:** Since the slope is given, we have dy/dx = 1- 1/x^{2}

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 = x^{2} + 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

(C) e (D) e^{2}

**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

(C) 4 (D) None of these

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

ax^{2} + 2hxy + by^{2} = 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

(C) 3 (D) 4

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

ax^{2} + by^{2} = 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

(C) d^{2}y/dx^{2}=0 (D) d^{2}y/dx^{2}=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 d^{2}y/dx^{2} = 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

(C) d^{2}y/dx^{2} =0 (D) d^{2}x/dy^{2} =0

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

Hence, the required differential equation will be given by d^{2}x/dy^{2} =0.

Hence (D) is the correct answer.

**30. The degree and order of the differential equation of all tangent lines to the parabola x ^{2} = 4y is:**

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

(C) 1, 3 (D) 1, 4

**Solution:** Equation of any tangent to x^{2} = 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) e^{2} (B) 2e^{2}

(C) 3e^{2} (D) 2e^{3}

**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} = e^{2} . e^{log2} = 2e^{2}.

Hence (B) is the correct answer.

**32. Equation of the curve passing through (3, 9) which satisfies the differential equation dy/dx = x+1/x ^{2} is:**

(A) 6xy = 3x^{2} – 6x + 29 (B) 6xy = 3x^{2} – 29x + 6

(C) 6xy = 3x^{3} – 29x – 6 (D) None of these

**Solution:** The given differential equation is

dy/dx = x +1/x^{2}

Integrating both sides we get, y= x^{2}/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= x^{2}/2 -1/x +29/6

Hence (C) is the correct answer.

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

(A) y^{2} = sinx (B) y = sin^{2}x

(C) y^{2} = cosx + 1 (D) None of these

**Solution:** On dividing the given equation by sin x,

2ydy/dx +y^{2}cot x = 2cos x

Now put y^{2} =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

⇒ y^{2} sin x = sin^{2}x + c

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

⇒ y^{2} = sin x.

Hence (A) is the correct answer.

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

(A) (B)

(C) (D) None of these

**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

(C) e^{x }(x – 1) + 1 = y (D) None of these

**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^{–y }dy = (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) d^{2}y/dx^{2} + 2dy/dx =0 =0 (B) d^{2}y/dx^{2} - 2dy/dx +2y =0

(C) d^{2}y/dx^{2} + 2dy/dx +2y =0 (D) d^{2}y/dx^{2} + 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)

⇒ d^{2}y/dx^{2} = e^{-x }(-A cosx -B sin x) –e^{-x} (-A sinx +B cos x)

Using (1) and (2), we get

d^{2}y/dx^{2} +dy/dx = -y-y-dy/dx

d^{2}y/dx^{2} +2dy/dx + 2y= 0.

Hence (C) is the correct answer.

**37. The solution of x ^{2} dy/dx –xy = 1+ cos y/x is:**

(A) tan y/2x = c-1/2x^{2 } (B) tan y/x = c+1/x^{ }

(C) cos y/x =1+c/x (D) x^{2 }=(c+x^{2}) tan y/x

**Solution:** Dividing the given equation by x^{2} throughout

dy/dx –y/x = 1/x^{2 }+ 1/x^{2} cos y/x .....(1)

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

Then equation (1) becomes v + x dv/dx –v =1/x^{2} +1/x^{2 }cos v

⇒ x^{3} dv/dx = 1+cos v

Then dv/ (1+cos v) = dx/x^{3}

Integrating both sides, ∫dv/ (1+cos v) = ∫dx/x^{3}

½ sec^{2}v/2 dv = -1/x +c

⇒ tan v/2 = -1/2x^{2} +c

Hence (A) is the correct answer.

**38. The order of differential equation d ^{2}y/dx^{2} = 1+dy/dx is:**

(A) 2 (B 3

(C) 1/2 (D) 6

**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

(C) parabolas (D) ellipses

**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

(C) cos x tan y = c (D) cos x sin 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+c_{1}

tan y/2 = e^{-cos x} e^{c}_{1}

⇒ 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

(C) y = e^{mx} + ce^{–ax} (D) (a + m) y = e^{mx} + ce^{–ax}

**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) +c_{1}

⇒ (a + m) y = e^{mx} + c_{1} (a + m) e^{–ax}

⇒ (a + m) y = e^{mx} + ce^{–ax}, where c = c_{1} (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

(C) sec x (D) sin 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

(C) Parabola whose vertex is at 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

(C) log (log x) (D) 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

(C) (x – y) e^{x+y} = 1 (D) (x – c) e^{x+y} + 1 = 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^{–y }dy/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 + A ^{3} of curves is represented by the differential equation of degree**

(A) 3 (B) 2

(C) 1 (D) None of these

**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 = x ^{3}; (x > 0) is :**

(A) x (B) log x

(C) –x (D) e^{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 – x ^{2}) dy + (2x^{2}y – y – ax^{3}) dx = 0 is e**

^{∫ Pdx}**, then P is:**

(A) (2x^{2} –ax^{3}) / (x (1-x^{2})) (B) 2x^{3} – 1

(C) (2x^{2} -1) / ax^{3} (D) (2x^{2}-1) / (x (1-x^{2}))

**Solution:** The given equation is x(1 – x^{2}) dy + (2x^{2}y – y – ax^{3}) dx = 0.

Hence ⇒ x(1 – x^{2}) dy/dx + 2x^{2}y – y – ax^{3} = 0.

⇒ dy/dx + {(2x^{2} -1)/ x (1-x^{2})} . y = ax^{3}/ x (1-x^{2})

Hence, clearly P is (2x^{3}-1)/ x (1-x^{2}).

Hence (D) is the correct answer.

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

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

(C) y = 4x (D) y + 4x + 1 = 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'= (x ^{2}+y^{2})/ (x^{2}-y^{2}), y (1) =1 has the slope at the point (1, 0) of the curve equal to:**

(A) -5/3 (B) –1

(C)1 (D) 5/3

**Solution:** dy/dx = (x^{2}+y^{2})/ (x^{2}-y^{2})

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

(C) y = 2x – 4 (D) y = 2x^{2} – 4

**Solutions **: (C)

**52. The solution of (2x-10y ^{3}) dx/dy + y=0 is:**

(A) x + y = ce^{2x} (B) y^{2} = 2x^{3} + c

(C) xy^{2} = 2y^{5} + c (D) x (y^{2} + xy) = 0

**Solution: **We can rewrite the given equation in the form** **

ydy/dx= -2x +10y^{3}

⇒ dx/dy +2x/y = 10y^{2}

This is a linear equation with P= 2/y and Q= 10y^{2 }

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

Hence, the solution is given by xy^{2} = ∫ 10 y^{2}.y^{2} dy +c

⇒ xy^{2}= 10y^{5}/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) log_{10}e (B) 2 log_{e} 10

(C) 2 log_{10} e (D) log_{10}e

**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

(C) Circle (D) parabola

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

⇒ ydy = xdx

Integrating both sides, y^{2}/2 –x^{2}/2 =c

⇒ y^{2} – x^{2} = 2c

which is a rectangular hyperbola

Hence (B) is the correct answer.

**55. The solution of the differential equation (1 – xy – x ^{5}y^{5}) dx – x^{2}(x^{4}y^{4} + 1)dy = 0 is given by**

(A) x = (B) x =

(C) x = (D) None of these

**Solution:** The given equation is dx – x (ydx+ xdy) = x^{5}y^{4 }(ydx + xdy)

⇒ dx/x = (1+x^{4}y^{4}) d (xy)

⇒ lnx = xy + 1/5 x^{5}y^{5} + ln c

⇒ x = c exp (xy+x^{5}y^{5}/5)

Hence (A) is the correct answer.

**56. If y= e ^{4x} +2e^{-x} satisfies the relation d^{3}y/dx^{3} + A dy/dx + By= 0 then value of A and B respectively are:**

(A) –13, 14 (B) –13, –12

(C) -13, 12^{ }(D) 12, –13

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

dy/dx = 4e^{4x} – 2e^{-x}

⇒ d^{2}y/dx^{2} = 16e^{4x} + 2e^{-x}

⇒ d^{3}y/dx^{3} = 64e^{4x }– 2e^{-x}

Putting these values in d^{3}y/dx^{3} + 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) (x^{2}-y^{2})dy/dx -2xy =0 (B) (x^{2}-y^{2})dy/dx+2xy =0

(C) (x^{2}-y^{2})dy/dx -xy =0 (D) (x^{2}-y^{2})dy/dx +xy =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 x^{2} + (y – a)^{2} = a^{2}

⇒ x^{2} + y^{2} – 2ay = 0 ....(1)

⇒ 2x+2ydy/dx -2ady/dx =0. .....(2)

Using (1) in (2), 2x+2ydy/dx – (x^{2}+y^{2})/y .dy/dx =0.

⇒ 2xy = (x^{2}+y^{2}-2y^{2})dy/dx

⇒ dy/dx = 2xy/ (x^{2}-y^{2})

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

(C) log x/ y = cx (D) log y/ x = cx

**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.