The highest order derivative present in the differential equation is and its order is 4.

The given differential equation is not a polynomial equation in derivatives as the term sin (y”’) is a T-function of derivative y”’. Therefore the degree is not defined.

Hence, order is 4 and degree is not defined.

**2. **Given: t’ + 5y = 0

The highest order derivative present in the differential equation is and its order is 1.

The given differential equation is a polynomial equation in derivative y’ and the highest power raised to highest order derivative y’ is one, so its degree is 1.

Hence, order is 1 and degree is 1.

The highest order derivative present in the differential equation is and its order is 2. The given differential equation is a polynomial equation in derivatives and the highest power raised to highest order derivative is one, so its degree is 1.

Hence, order is 2 and degree is 1.

The highest order derivative present in the differential equation is and its order is 2.

The given differential equation is not a polynomial equation in derivatives as the term is a T-function of derivative . Therefore the degree is not defined.

Hence, order is 2 and degree is not defined.

The highest order derivative present in the differential equation is and its order is 2.

The given differential equation is a polynomial equation in derivatives and the highest power raised to highest order is one, so its degree is 1.

Hence, order is 2 and degree is 1.

**6.** Given: (y”’)^{2} + (y”)^{3} + (y’)^{4} + y^{5} = 0

The highest order derivative present in the differential equation is y”’ and its order is 3.

The given differential equation is a polynomial equation in derivatives and the highest power raised to highest order y”’ is two, so its degree is 2.

Hence, order is 3 and degree is 2.

**7.** Given: y”’ + 2y” + y’ = 0

The highest order derivative present in the differential equation is y”’ and its order is 3.

The given differential equation is a polynomial equation in derivatives y”’, y” and y’ and the highest power raised to highest order y”’ is two, so its degree is 1.

Hence, order is 3 and degree is 1.

**8.** Given: y’ + y = e^{x}

The highest order derivative present in the differential equation is y’ and its order is 1.

The given differential equation is a polynomial equation in derivative y’. It may be noted that e^{x }is an exponential function and not a polynomial function but is not an exponential function of derivatives and the highest power raised to highest order derivative y’ is one so its degree is one.

Hence, order is 1 and degree is 1.

**9.** y” + (y’)^{2} + 2y = 0

The highest order derivative present in the differential equation is y” and its order is 2.

The given differential equation is a polynomial equation in derivatives y” and y’ and the highest power raised to highest order y” is one, so its degree is 1.

Hence, order is 2 and degree is 1.

**10.** y” + 2y’ + sin y = 0

The highest order derivative present in the differential equation is y” and its order is 2.

The given differential equation is a polynomial equation in derivative y” and y’. It may be noted that sin y is not a polynomial function of y, it is a T-function of y but is not a T-function of derivatives and the highest power raised to highest order derivative y” is one so its degree is one.

Hence, order is 2 and degree is 1.

This equation is not a polynomial in derivatives as is a T-function of derivative

Therefore, degree of given equation is not defined.

Hence, option (D) is correct.

The highest order derivative present in the differential equation is and its order is 2.

Therefore, option (A) is correct.

**13.** Given: y = e^{x} + 1 : y” – y’ = 0

To prove: y is a solution of the differential equation y” – y’ = 0 ……….(ii)

Proof: From eq. (i), y’ = e^{x} + 0 = e^{x} and y” = e^{x}

∴ L.H.S. of eq. (ii), y” – y’ = e^{x} – e^{x} = 0 = R.H.S.

Hence, given by eq. (i) is a solution of y” – y’ = 0.

**14.** Given: ** **…..(i)

To prove: y is a solution of the differential equation y’ – 2x – 2 = 0 ……….(ii)

Proof: From, eq. (i),

y’ = 2x + 2

L.H.S. of eq. (ii),

= y’ – 2x – 2

= (2x + 2) – 2x – 2

= 2x + 2 – 2x – 2 = 0 = R.H.S.

Hence, y given by eq. (i) is a solution of y’ – 2x – 2 = 0.

**15.** Given: y = cos x + C …(i)

To prove: y is a solution of the differential equation y’ + sin x = 0 ……….(ii)

Proof: From eq. (i),

y’ = -sin x

L.H.S. of eq. (ii),

y’ + sin x = -sin x + sin x = 0 = R.H.S.

Hence, y given by eq. (i) is a solution of y’ + sin x = 0.

**17. **Given: y = Ax ……….(i)

To prove: given by eq. (i) is a solution of differential equation xy’ = y(x ≠ 0) …(ii)

Proof: From eq. (i)

y’ = A(1) = A

L.H.S. of eq. (ii)

= xy’ = xA = Ax = y = R.H.S. of eq. (ii)

∴ y given by eq. (i) is a solution of differential equation xy’ = y(x ≠ 0).

**18.** Given: y = x sin x …(i)

**19.** Given: xy = log y + C ….(i)

**20.** Given: y – cos y = x …..(i)

To prove: y given by eq. (i) is a solution of differential equation

(y sin y + cos y + x)y’ = y ….(ii)

Proof: Differentiating both sides of eq. (i) w.r.t x, we have

Putting the value of x from eq. (i) and value of y’ from eq. (iii) in L.H.S. of eq. (ii),

Hence, Function given by eq. (i) is a solution of (y sin y + cos y + x)y’ = y.

**21.** Given: x + y = tan^{-1} y …..(i)

To prove: given by eq. (i) is a solution of differential equation y^{2} y’ + y^{2} + 1 = 0 …(ii)

Proof: Differentiating both sides of eq. (i) w.r.t x we have

Hence, Function given by eq. (i) is a solution of y^{2} y’ + y^{2} + 1 = 0.

**23. **Option (D) is correct.

The number of arbitrary constants (c_{1}, c_{2}, c_{3} etc.) in the general solution of a differential equation of nth order is n.

**24. **The number of arbitrary constants in a particular solution of a differential equation of any order is zero (0) as a particular solution is a solution which contains no arbitrary constant.

Therefore, option (D) is correct.

**25. ** Given: Equation of the family of curves

Here there are two arbitrary constants a and b, therefore we will differentiate both sides two time w.r.t.x,

**26. **Given: Equation of the family of curves y^{2} = a(b^{2} – x)^{2} ……….(i)

Here there are two arbitrary constants and therefore we will differentiate both sides two times w.r.t.x,

**27. **Given: Equation of the family of curves y = ae^{3x} + be^{-2x }……….(i)

Here there are two arbitrary constants and therefore we will differentiate both sides two times w.r.t.x,

**28. **Given: Equation of the family of curves y = e^{2x} (a + bx) ……….(i)

Here there are two arbitrary constants and therefore we will differentiate both sides two times w.r.t.x,

**29. **Given: Equation of the family of curves y = e^{x }(a cos x + b sin x) ……….(i)

Here there are two arbitrary constants and therefore we will differentiate both sides two times w.r.t.x

**30. **It is clear that if a circle touches axis at the origin must have its centre on axis, because axis being at right angles to axis is the normal or line of radius of the circle.

Therefore, the centre of the circle is (r, 0) where is the radius of the circle.

∴ Equation of the required circle is (x – r)^{2} + (y – 0)^{2} = r^{2}

⇒ x^{2} + r^{2} + 2rx + y^{2} = r^{2}

⇒ x^{2} + y^{2} = 2rx ….(i)

Here r is the only arbitrary constant.

∴ Differentiating w.r.t.x, we get

**31.** We know that equation of parabola having vertex at origin and axis along positive y-axis is x^{2} = 4ay

Here a is the only arbitrary constant. Therefore differentiating w.r.t. x, we get

**32. ** We know that equation of ellipse having foci on axis i.e., vertical ellipse with major axis as y-axis is

Here there are two arbitrary constants a and b, therefore we will differentiate both sides two times w.r.t. x,

**33.** We know that equation of hyperbolas having foci on x-axis and centre at origin is

Here there are two arbitrary constants a and b, therefore we will differentiate both sides two times w.r.t. x

**34. ** We know that on y-axis, x = 0

∴ Centre of the circle on y-axis is (0, β).

∴ Equation of the circle having centre on y-axis and radius β unit is

(x – 0)^{2} + (y – β)^{2} = (3)^{2}

⇒ x^{2} + (y – β)^{2} = 9 …..(i)

Here β is the only arbitrary constant, therefore we will differentiate only once.

**36.** Given: y = x

On putting these values in the given option, we get the correct answer in option (C).

L.H.S. of differential equation of option

= 0 – x^{2}(1) + x(x)

= -x^{2} + x^{2} = 0

= R.H.S. of option (C)

Therefore, option (C) is correct.

**43.** Given: Differential equation y log y dx – x dy = 0

[If all the terms in the solution of a differential equation involve log, it is better to use log c or log|c| instead of c in the solution.]

⇒ |t| = |xc| ⇒ t = ±xc

⇒ log y = ±xc = ax where a = ±c

⇒ y = e^{ax}

**46. ** Given: Differential equation e^{x} tan y dx + (1 – e^{x}) sec^{2} y dy = 0

Dividing each term by (1 – e^{x})tan y, we get

Comparing the coefficients of x^{2} on both sides, A + B = 2 ……….(iii)

Comparing the coefficients of x on both sides, B + C = 1 ……….(iv)

Comparing constants on both sides, A + C = 0 ……….(v)

From eq. (iii) – (iv), we have A – C = 1 ……….(vi)

Adding eq. (v) and (vi), we have 2A = 1

Comparing the coefficients of x^{2} on both sides, A + B + C = 0 ……….(iii)

Comparing the coefficients of x on both sides, B + C = 0

⇒ C = B ……….(iv)

Comparing constants on both sides, – A = 1

⇒ A = – 1 ……….(v)

Putting A = -1 and C = B in eq. (iii),

Putting the values of A, B and C in eq. (ii), we get

Now putting y = 1 and x = 0 in eq. (i), we get 1 = C sec 0 = C

Putting C = 1 in eq. (i), we get the required general solution

⇒ y = sec x

**51. ** Given: Differential equation y’ = e^{x} sin x

Now putting x = 1, y = -1 in eq.(i),

-1 -1 = log(1) + c

⇒ c = -2

Putting this value of c in eq.(i) to get the required solution curve

y – x = log(y + 2)^{2} .(x)^{2} – 2

⇒ y – x + 2 = log(y + 2)^{2} .(x)^{2}

**53.** Let P(x, y) be any point on the required curve.

According to the question, Slope of the tangent to the curve at P(x, y) × y = x

Now it is given that curve y^{2} = x^{2} + C passes through the point (0, -2).

Therefore, putting x = 0 and y = -2 in this equation, we get C = 4

Putting the value of C in the equation y^{2} = x^{2} + C,

y^{2} = x^{2} + 4

⇒ y^{2} – x^{2} = 4.

**54. **According to the question, slope of the tangent at any point P(x, y) of the required curve

= 2. Slope of the line joining the point of contact P(x, y) to the given point A(-4, -3).

Now it is given that curve (i) passes through the point (-2, 1).

Therefore, putting x = -2 and y = 1 in eq.(i),

⇒ 1 + 3 = C(-2 + 4)^{2}

⇒ 4 = 4C

⇒ C = 1

Putting C = 1 in eq.(i), we get the required solution,

y + 3 = (x + 4)^{2}

⇒ (x + 4)^{2} = y + 3.

**55. **Let x be the radius of the spherical balloon at time t.

Given: Rate of change of volume of spherical balloon is constant = k (say)

Now it is given that initially radius is 3 units, when t = 0, x = 3.

Therefore, putting t = 0, x = 3 in eq. (i), 4π(27) = c

⇒ c = 36π ……….(ii)

Again when t = 3 sec, then x = 6 units

Therefore, putting t = 3 and x = 6 in eq.(i),

**56. **Let P be the principal (amount) at the end of t years.

According to the given condition, rate of increase of principal per year = r% (of principal)

**57. **Let P be the principal (amount) at the end of t years.

According to the given condition, rate of increase of principal per year = 5% (of principal)

**58. **Let x be the bacteria present in the culture at time t hours.

According to the question,

Rate of growth of bacteria is proportional to the number present

Now it is given that initially the bacteria count is x_{0} (say) = 1,00,000

⇒ when t = 0, then x = x_{0}

Putting these values in eq. (i) log x_{0} = c

Putting log x_{0} = c in eq. (i), we get

Now it is given also that the number of bacteria increased by 10% in 2 hours.

Therefore, increase in bacteria in 2 hours

∴ x, the amount of bacteria at t = 2 = 1,00,000 + 10,000 = 1,10,000 = x_{1} (say)

Putting x = x_{1} and t = 2 in eq.(ii),

**60.** Given: Differential equation (x^{2} + xy) dy = (x^{2} + y^{2}) dx …..(i)

Here degree of each coefficients of and is same therefore, it is homogenous.

**62.** Given: Differential equation (x – y) dy + (x + y) dx = 0 ……….(i)

This given equation is homogeneous because each coefficients of dx and dy is of degree 1.

**63.** Given: Differential equation: (x^{2} – y^{2}) dx + 2xy dy = 0

This equation is homogeneous because degree of each coefficient of dx and dy is same i.e., 2

Therefore, the given differential equation is homogeneous as all terms of x and y are of same degree i.e., degree 2.

**70.** Given: Differential equation (x + y) dy + (x – y) dx = 0; y = 1 when x = 1 …..(i)

Therefore the given differential equation is homogeneous because each coefficient of dx and dy is same i.e., degree 1.

**71.** Given: Differential equation x^{2} dy + (xy + y^{2}) dx = 0; y = 1 when x = 1

Therefore the given differential equation is homogeneous.

Therefore the given differential equation is homogeneous because each coefficient of dx and dy is same i.e., degree 2.

**75. **We know that a homogeneous differential equation of the form can be solved by the substitution

Therefore, option (C) is correct.

**76. ** D

**92.** Slope of the tangent to the curve at any point (x, y) = Sum of coordinates of the point (x, y)

Applying Product rule of Integration,

Now, since curve (i) passes through the origin (0, 0), therefore putting x = 0, y = 0 in eq. (i)

⇒ -c = -1

⇒ c = 1

Putting c = 1 in eq. (i),

y = -x – 1 + e^{x}

⇒ y + x + 1 = e^{x}

**93. **According to the question, Sum of the coordinates of any point say (x, y) on the curve

= Magnitude of the slope of the tangent to the curve + 5

Now, since curve (i) passes through the origin (0, 2), therefore putting x = 0, y = 2 in eq. (i)

⇒ 0 + 2 = 4 + ce^{0}

⇒ c = -2

Putting c = -2 in eq. (i),

⇒ x + y = 4 – 2e^{x}

⇒ y = 4 – x – 2e^{x}

**96. **(i) Given: Differential equation

The highest order derivative present in this differential equation is and hence order of this differential equation if 2.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative is 1.

Therefore, Order = 2, Degree = 1

(ii) Given: Differential equation

The highest order derivative present in this differential equation is and hence order of this differential equation if 1.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative is 3.

Therefore, Order = 1, Degree = 3

(iii) Given: Differential equation

The highest order derivative present in this differential equation is and hence order of this differential equation if 4.

The given differential equation is not a polynomial equation in derivatives therefore, degree of this differential equation is not defined.

Therefore, Order = 4, Degree not defined

**97.** (i) The given function is xy = ae^{x} + be^{-x} + x^{2} ……….(i)

Therefore, Function given by eq. (i) is a solution of D.E. (ii).

(ii) The given function is y = e^{x} (a cos x + b sin x) ……….(i)

Therefore, Function given by eq. (i) is a solution of D.E. (ii).

(iii) The given function is y = x sin 3x ………(i)

Therefore, Function given by eq. (i) is a solution of D.E. (ii).

(iv) The given function is x^{2} = 2y^{2} log y ……….(i)

Therefore, Function given by eq. (i) is a solution of D.E. (ii).

**98.** Equation of the given family of curves is (x – a)^{2} + 2y^{2} = a^{2}

⇒ x^{2} + a^{2} – 2ax + 2y^{2} = a^{2}

⇒ x^{2} – 2ax + 2y^{2} = 0

⇒ x^{2} + 2y^{2} = 2ax ….(i)

Here number of arbitrary constants is one only (a).

So, we will differentiate both sides of equation only once, w.r.t.x,

Dividing eq.(i) by eq.(ii), we have

**99.** Given: Differential equation (x^{3} – 3xy^{2}) dx = (y^{3} – 3x^{2}y) dy …..(i)

Here each coefficient of dx and dy is of same degree, i.e., 3, therefore differential equation looks to be homogeneous.

Therefore, the given differential equation is homogeneous.

Comparing coefficients of like powers of

ν^{3} A – B – C = 1 ……….(v)

ν^{2} A + B – D = 0 ……….(vi)

ν A – B + C = –3 ……….(vii)

Constants A + B + D = 0 ………(viii)

Now eq. (v) – eq. (vii)

⇒ – 2C = 4 ⇒ C = – 2

Eq. (vi) – eq. (viii)

⇒ – 2D = 0 ⇒ D = 0

Putting C = –2 in eq. (v), A – B + 2 = 1

⇒ A – B = –1 ……….(ix)

Putting D = 0 in eq. (vi) A + B = 0 ……….(x)

Adding eq. (ix) and (x) 2A = –1

**100.** We know that the circle in the first quadrant which touches the co-ordinates axes has centre (a, a) where is the radius of the circle.

∴ Equation of the circle is

(x – a) + (y – a) = a ….(i)

⇒ x + y – 2ax – 2ay + a = 0

Differentiating with respect to x,

Multiplying every term in the numerator and denominator of L.H.S. by 3, and dividing every term by 2√3.

**103.** Given: Differential equation sin x cos y dx + cos x sin y dy = 0

**104.** Given: Differential equation (1 + e^{2x}) dy + (1 + y^{2})e^{x} dx = 0

Dividing every term by (1 + y^{2}) (1 + e^{2x}), we have

**105.** Given: Differential equation ye^{x/y} dx = (xe^{x/y} + y^{2}) dy (y ≠ 0)

It is not a homogeneous differential equation because of presence of only as a factor, yet it can be solved by putting

Putting these values in eq. (i), we get

**106.** Given: Differential equation (x – y) (dx + dy) = dx – dy

**110. **Let P be the population of the village at time t.

According to the question, Rate of increase of population of the village is proportional to the number of inhabitants.

Now Population of the village was P = 20,000 in the year 1999.

Let us take the base year 1999 as t = 0.

Putting t = 0 and P = 20000 in eq. (i), log 20000 = c

Now putting log 2000 = c in eq. (i), log P = kt + log 20000

⇒ log P – log 20000 = kt