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NCERT Solutions · Class 12 Maths · NCERT Mathematics Part I and II Part II · Chapter 9

Chapter 9: Differential Equations (Calculus)

Step-by-step solutions to all exercises of Chapter 9, Differential Equations (NCERT Class 12 Maths Part II, 2026-27): Exercise 9.1 (order and degree), 9.2 (verifying solutions), 9.3 (variables separable), 9.4 (homogeneous equations), 9.5 (linear equations) and the Miscellaneous Exercise. All 98 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 9 (PDF)

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Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-12-maths/chapter-9-differential-equations

Order = highest derivative; degree = its power when the equation is a polynomial in the derivatives (else not defined). Variables separable: bring all y to one side, all x to the other, integrate. Homogeneous $\frac{d y}{d x} = F (\frac{y}{x})$ : put $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Linear $\frac{d y}{d x} + P y = Q$ : integrating factor $IF = e^{\int P d x}$ , solution $y \cdot IF = \int Q \cdot IF d x + C$ .

Exercise 9.1

Determine order and degree (if defined):

\frac{d ^{4} y}{d x ^{4}} + sin (y^{'''}) = 0

Solution

Highest derivative: $y^{''''}$ , so order 4. The term $sin (y^{'''})$ is not a polynomial in the derivatives, so the degree is not defined.

Order 4; degree not defined

Determine order and degree: y′ + 5y = 0

Solution

Highest derivative y′ with power 1.

Order 1; degree 1

Determine order and degree:

(\frac{d s}{d t})^{4} + 3 s \frac{d ^{2} s}{d t ^{2}} = 0

Solution

Highest derivative is $\frac{d ^{2} s}{d t ^{2}}$ (order 2), appearing with power 1.

Order 2; degree 1

Determine order and degree:

(\frac{d ^{2} y}{d x ^{2}})^{2} + cos (\frac{d y}{d x}) = 0

Solution

Order 2. $cos (\frac{d y}{d x})$ is not a polynomial term.

Order 2; degree not defined

Determine order and degree:

\frac{d ^{2} y}{d x ^{2}} = cos 3 x + sin 3 x

Solution

Order 2; degree 1

Determine order and degree: (y‴)² + (y″)³ + (y′)⁴ + y⁵ = 0

Solution

Highest derivative y‴, power 2.

Order 3; degree 2

Determine order and degree: y‴ + 2y″ + y′ = 0

Solution

Order 3; degree 1

Determine order and degree: y′ + y = eˣ

Solution

Order 1; degree 1

Determine order and degree: y″ + (y′)² + 2y = 0

Solution

Highest derivative y″ appears with power 1.

Order 2; degree 1

Determine order and degree: y″ + 2y′ + sin y = 0

Solution

$sin y$ is a function of y, not of a derivative, so the equation is a polynomial in the derivatives.

Order 2; degree 1

The degree of the differential equation

(\frac{d ^{2} y}{d x ^{2}})^{3} + (\frac{d y}{d x})^{2} + sin (\frac{d y}{d x}) + 1 = 0

is (A) 3 (B) 2 (C) 1 (D) not defined

Solution

$sin (\frac{d y}{d x})$ makes it a non-polynomial in the derivatives.

(D)

The order of the differential equation

2 x^{2} \frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + y = 0

is (A) 2 (B) 1 (C) 0 (D) not defined

Solution

(A)

Exercise 9.2

Verify that y = eˣ + 1 is a solution of y″ – y′ = 0.

Solution

$y^{'} = e^{x}$ , $y^{''} = e^{x}$ . So $y^{''} - y^{'} = 0$ .

Verified.

Verify that y = x² + 2x + C is a solution of y′ – 2x – 2 = 0.

Solution

$y^{'} = 2 x + 2$ , so $y^{'} - 2 x - 2 = 0$ .

Verified.

Verify that y = cos x + C is a solution of y′ + sin x = 0.

Solution

$y^{'} = - sin x$ , so $y^{'} + sin x = 0$ .

Verified.

Verify that

y = 1 + x^{2}

is a solution of

y^{'} = \frac{x y}{1 + x ^{2}}

Solution

$y^{'} = \frac{x}{1 + x ^{2}} = \frac{x 1 + x ^{2}}{1 + x ^{2}} = \frac{x y}{1 + x ^{2}}$ .

Verified.

Verify that y = Ax is a solution of xy′ = y (x ≠ 0).

Solution

$y^{'} = A$ , so $x y^{'} = A x = y$ .

Verified.

Verify that y = x sin x is a solution of

x y^{'} = y + x x^{2} - y^{2}

(x ≠ 0 and x > y or x < –y).

Solution

$y^{'} = sin x + x cos x$ , so $x y^{'} = x sin x + x^{2} cos x = y + x^{2} cos x$ .

RHS: $x x^{2} - x^{2} sin^{2} x = x \cdot x cos x = x^{2} cos x$ (in the given range). So LHS = RHS.

Verified.

Verify that xy = log y + C is a solution of

y^{'} = \frac{y ^{2}}{1 - x y}

(xy ≠ 1).

Solution

Differentiate: $y + x y^{'} = \frac{y ^{'}}{y} \Rightarrow y^{'} (\frac{1}{y} - x) = y \Rightarrow y^{'} \cdot \frac{1 - x y}{y} = y$ .

$y^{'} = \frac{y ^{2}}{1 - x y}$ : verified.

Verify that y – cos y = x is a solution of (y sin y + cos y + x) y′ = y.

Solution

Differentiate: $y^{'} + sin y \cdot y^{'} = 1 \Rightarrow y^{'} = \frac{1}{1 + sin y}$ .

Using $x = y - cos y$ : $y sin y + cos y + x = y sin y + y = y (1 + sin y)$ .

LHS $= y (1 + sin y) \cdot \frac{1}{1 + sin y} = y$ .

Verified.

Verify that x + y = tan⁻¹ y is a solution of y² y′ + y² + 1 = 0.

Solution

Differentiate: $1 + y^{'} = \frac{y ^{'}}{1 + y ^{2}} \Rightarrow y^{'} (\frac{1}{1 + y ^{2}} - 1) = 1 \Rightarrow y^{'} \cdot \frac{- y ^{2}}{1 + y ^{2}} = 1$ .

So $- y^{2} y^{'} = 1 + y^{2}$ , i.e. $y^{2} y^{'} + y^{2} + 1 = 0$ .

Verified.

Verify that

y = a^{2} - x^{2}

, x ∈ (–a, a) is a solution of

x + y \frac{d y}{d x} = 0

(y ≠ 0).

Solution

$\frac{d y}{d x} = \frac{- x}{a ^{2} - x ^{2}} = - \frac{x}{y}$ , so $x + y \frac{d y}{d x} = x - x = 0$ .

Verified.

The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) 0 (B) 2 (C) 3 (D) 4

Solution

The general solution has as many arbitrary constants as the order.

(D)

The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) 3 (B) 2 (C) 1 (D) 0

Solution

A particular solution has no arbitrary constants.

(D)

Exercise 9.3

Find the general solution:

\frac{d y}{d x} = \frac{1 - cos x}{1 + cos x}

Solution

$\frac{1 - cos x}{1 + cos x} = tan^{2} \frac{x}{2} = sec^{2} \frac{x}{2} - 1$ , so $y = \int (sec^{2} \frac{x}{2} - 1) d x$ .

$y = 2 tan \frac{x}{2} - x + C$

Find the general solution:

\frac{d y}{d x} = 4 - y^{2}

(–2 < y < 2)

Solution

$\frac{d y}{4 - y ^{2}} = d x \Rightarrow sin^{- 1} \frac{y}{2} = x + C$ .

$y = 2 sin (x + C)$

Find the general solution:

\frac{d y}{d x} + y = 1

(y ≠ 1)

Solution

$\frac{d y}{1 - y} = d x \Rightarrow - lo g ∣1 - y ∣ = x + C \Rightarrow 1 - y = \pm e^{- C} e^{- x}$ .

$y = 1 + A e^{- x}$

Find the general solution: sec² x tan y dx + sec² y tan x dy = 0

Solution

Divide by $tan x tan y$ : $\frac{sec ^{2} x}{tan x} d x + \frac{sec ^{2} y}{tan y} d y = 0$ .

$lo g ∣ tan x ∣ + lo g ∣ tan y ∣ = lo g C$ .

$tan x tan y = C$

Find the general solution: (eˣ + e⁻ˣ) dy – (eˣ – e⁻ˣ) dx = 0

Solution

$d y = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} d x$ ; the numerator is the derivative of the denominator.

$y = lo g (e^{x} + e^{- x}) + C$

Find the general solution:

\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})

Solution

$\frac{d y}{1 + y ^{2}} = (1 + x^{2}) d x$ .

$tan^{- 1} y = x + \frac{x ^{3}}{3} + C$

Find the general solution: y log y dx – x dy = 0

Solution

$\frac{d y}{y lo g y} = \frac{d x}{x} \Rightarrow lo g ∣ lo g y ∣ = lo g ∣ x ∣ + lo g c \Rightarrow lo g y = c x$ .

$y = e^{c x}$

Find the general solution:

x^{5} \frac{d y}{d x} = - y^{5}

Solution

$y^{- 5} d y = - x^{- 5} d x \Rightarrow \frac{y ^{- 4}}{- 4} = - \frac{x ^{- 4}}{- 4} + C^{'} \Rightarrow - y^{- 4} = x^{- 4} + 4 C^{'}$ .

$x^{- 4} + y^{- 4} = C$

Find the general solution:

\frac{d y}{d x} = sin^{- 1} x

Solution

$y = \int sin^{- 1} x d x = x sin^{- 1} x - \int \frac{x}{1 - x ^{2}} d x = x sin^{- 1} x + 1 - x^{2}$ .

$y = x sin^{- 1} x + 1 - x^{2} + C$

Find the general solution: eˣ tan y dx + (1 – eˣ) sec² y dy = 0

Solution

$\frac{sec ^{2} y}{tan y} d y = \frac{e ^{x}}{e ^{x} - 1} d x \Rightarrow lo g ∣ tan y ∣ = lo g ∣ e^{x} - 1∣ + lo g C$ .

$tan y = C (1 - e^{x})$ (sign absorbed in C)

Find a particular solution:

(x^{3} + x^{2} + x + 1) \frac{d y}{d x} = 2 x^{2} + x

; y = 1 when x = 0

Solution

$x^{3} + x^{2} + x + 1 = (x + 1) (x^{2} + 1)$ . $\frac{2 x ^{2} + x}{( x + 1 ) ( x ^{2} + 1 )} = \frac{1/2}{x + 1} + \frac{\frac{3}{2} x - \frac{1}{2}}{x ^{2} + 1}$ .

$y = \frac{1}{2} lo g ∣ x + 1∣ + \frac{3}{4} lo g (x^{2} + 1) - \frac{1}{2} tan^{- 1} x + C$ . At $x = 0$ , $y = 1$ : $C = 1$ .

$y = \frac{1}{4} lo g [(x + 1)^{2} (x^{2} + 1)^{3}] - \frac{1}{2} tan^{- 1} x + 1$

Find a particular solution:

x (x^{2} - 1) \frac{d y}{d x} = 1

; y = 0 when x = 2

Solution

$\frac{1}{x ( x - 1 ) ( x + 1 )} = - \frac{1}{x} + \frac{1/2}{x - 1} + \frac{1/2}{x + 1}$ , so $y = \frac{1}{2} lo g \frac{x ^{2} - 1}{x ^{2}} + C$ .

At $x = 2$ : $0 = \frac{1}{2} lo g \frac{3}{4} + C \Rightarrow C = - \frac{1}{2} lo g \frac{3}{4}$ .

$y = \frac{1}{2} lo g (\frac{x ^{2} - 1}{x ^{2}}) - \frac{1}{2} lo g \frac{3}{4}$

Find a particular solution:

cos (\frac{d y}{d x}) = a

(a ∈ R); y = 1 when x = 0

Solution

$\frac{d y}{d x} = cos^{- 1} a \Rightarrow y = x cos^{- 1} a + C$ . At $x = 0$ : $C = 1$ . So $\frac{y - 1}{x} = cos^{- 1} a$ .

$cos (\frac{y - 1}{x}) = a$

Find a particular solution:

\frac{d y}{d x} = y tan x

; y = 1 when x = 0

Solution

$\frac{d y}{y} = tan x d x \Rightarrow lo g ∣ y ∣ = lo g ∣ sec x ∣ + C$ . At $x = 0$ , $y = 1$ : $C = 0$ .

$y = sec x$

Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = eˣ sin x.

Solution

$y = \int e^{x} sin x d x = \frac{e ^{x} ( sin x - cos x )}{2} + C$ (by parts twice). At (0, 0): $0 = - \frac{1}{2} + C \Rightarrow C = \frac{1}{2}$ .

$2 y - 1 = e^{x} (sin x - cos x)$

For the differential equation

x y \frac{d y}{d x} = (x + 2) (y + 2)

, find the solution curve passing through the point (1, –1).

Solution

$\frac{y}{y + 2} d y = \frac{x + 2}{x} d x \Rightarrow (1 - \frac{2}{y + 2}) d y = (1 + \frac{2}{x}) d x$ .

$y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ + C$ . At (1, −1): $- 1 = 1 + C \Rightarrow C = - 2$ .

$y - x + 2 = lo g (x^{2} (y + 2)^{2})$

Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Solution

$y \frac{d y}{d x} = x \Rightarrow y d y = x d x \Rightarrow \frac{y ^{2}}{2} = \frac{x ^{2}}{2} + C^{'}$ , so $y^{2} - x^{2} = C$ . At (0, −2): $C = 4$ .

$y^{2} - x^{2} = 4$

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Solution

$\frac{d y}{d x} = \frac{2 ( y + 3 )}{x + 4} \Rightarrow \frac{d y}{y + 3} = \frac{2 d x}{x + 4} \Rightarrow y + 3 = K (x + 4)^{2}$ .

At (−2, 1): $4 = 4 K \Rightarrow K = 1$ .

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

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Solution

$\frac{d V}{d t} = k \Rightarrow \frac{4}{3} π r^{3} = k t + C$ .

$t = 0$ , $r = 3$ : $C = 36 π$ . $t = 3$ , $r = 6$ : $288 π = 3 k + 36 π \Rightarrow k = 84 π$ .

$\frac{4}{3} π r^{3} = 84 π t + 36 π \Rightarrow r^{3} = 63 t + 27$ .

$r = (63 t + 27)^{1/3}$

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (logₑ2 = 0.6931).

Solution

$\frac{d P}{d t} = \frac{r}{100} P \Rightarrow P = P_{0} e^{r t /100}$ . Doubling in 10 years: $2 = e^{r /10} \Rightarrow r = 10 lo g_{e} 2 = 6.931$ .

r = 6.93%

In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e⁰·⁵ = 1.648).

Solution

$P = 1000 e^{0.05 \times 10} = 1000 e^{0.5} = 1000 \times 1.648$ .

₹1648

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solution

$\frac{d N}{d t} = k N \Rightarrow N = N_{0} e^{k t}$ . After 2 h: $1, 10, 000 = 1, 00, 000 e^{2 k} \Rightarrow e^{2 k} = \frac{11}{10}$ , $k = \frac{1}{2} lo g \frac{11}{10}$ .

For $N = 2, 00, 000$ : $2 = e^{k t} \Rightarrow t = \frac{lo g 2}{k} = \frac{2 lo g 2}{lo g \frac{11}{10}}$ .

$\frac{2 lo g 2}{lo g ( 11/10 )}$ hours (about 14.5 h)

The general solution of the differential equation

\frac{d y}{d x} = e^{x + y}

is (A) eˣ + e⁻ʸ = C (B) eˣ + eʸ = C (C) e⁻ˣ + eʸ = C (D) e⁻ˣ + e⁻ʸ = C

Solution

$e^{- y} d y = e^{x} d x \Rightarrow - e^{- y} = e^{x} + c \Rightarrow e^{x} + e^{- y} = C$ .

(A)

Exercise 9.4

Show that the equation is homogeneous and solve: (x² + xy) dy = (x² + y²) dx

Solution

$\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x ^{2} + x y}$ (degree 0 in x, y: homogeneous). Put $y = v x$ :

$v + x \frac{d v}{d x} = \frac{1 + v ^{2}}{1 + v} \Rightarrow x \frac{d v}{d x} = \frac{1 - v}{1 + v}$ .

$\frac{1 + v}{1 - v} d v = \frac{d x}{x}$ , and $\frac{1 + v}{1 - v} = - 1 + \frac{2}{1 - v}$ : $- v - 2 lo g ∣1 - v ∣ = lo g ∣ x ∣ + C$ .

With $v = \frac{y}{x}$ : $lo g ∣ x ∣ - 2 lo g ∣ x - y ∣ = C + \frac{y}{x}$ , i.e. $\frac{x}{( x - y ) ^{2}} = C_{1} e^{y / x}$ .

$(x - y)^{2} = C x e^{- y / x}$

Show that the equation is homogeneous and solve:

y^{'} = \frac{x + y}{x}

Solution

$v + x \frac{d v}{d x} = 1 + v \Rightarrow d v = \frac{d x}{x} \Rightarrow v = lo g ∣ x ∣ + C$ .

$y = x lo g ∣ x ∣ + C x$

Show that the equation is homogeneous and solve: (x – y) dy – (x + y) dx = 0

Solution

$\frac{d y}{d x} = \frac{x + y}{x - y}$ ; $y = v x$ : $x \frac{d v}{d x} = \frac{1 + v}{1 - v} - v = \frac{1 + v ^{2}}{1 - v}$ .

$\frac{1 - v}{1 + v ^{2}} d v = \frac{d x}{x} \Rightarrow tan^{- 1} v - \frac{1}{2} lo g (1 + v^{2}) = lo g ∣ x ∣ + C$ .

Since $\frac{1}{2} lo g (1 + v^{2}) + lo g ∣ x ∣ = \frac{1}{2} lo g (x^{2} + y^{2})$ :

$tan^{- 1} (\frac{y}{x}) = \frac{1}{2} lo g (x^{2} + y^{2}) + C$

Show that the equation is homogeneous and solve: (x² – y²) dx + 2xy dy = 0

Solution

$\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$ ; $y = v x$ : $x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} - v = - \frac{1 + v ^{2}}{2 v}$ .

$\frac{2 v}{1 + v ^{2}} d v = - \frac{d x}{x} \Rightarrow lo g (1 + v^{2}) = - lo g ∣ x ∣ + lo g C \Rightarrow x (1 + v^{2}) = C$ .

$x^{2} + y^{2} = C x$

Show that the equation is homogeneous and solve:

x^{2} \frac{d y}{d x} = x^{2} - 2 y^{2} + x y

Solution

$\frac{d y}{d x} = 1 - 2 v^{2} + v$ ; $y = v x$ : $x \frac{d v}{d x} = 1 - 2 v^{2} \Rightarrow \frac{d v}{1 - 2 v ^{2}} = \frac{d x}{x}$ .

$\int \frac{d v}{1 - ( 2 v ) ^{2}} = \frac{1}{2 2} lo g \frac{1 + 2 v}{1 - 2 v}$ .

$\frac{1}{2 2} lo g \frac{x + 2 y}{x - 2 y} = lo g ∣ x ∣ + C$

Show that the equation is homogeneous and solve:

x d y - y d x = x^{2} + y^{2} d x

Solution

$\frac{d y}{d x} = \frac{y + x ^{2} + y ^{2}}{x}$ ; $y = v x$ : $x \frac{d v}{d x} = 1 + v^{2}$ .

$\frac{d v}{1 + v ^{2}} = \frac{d x}{x} \Rightarrow lo g v + 1 + v^{2} = lo g ∣ x ∣ + lo g C \Rightarrow v + 1 + v^{2} = C x$ .

$y + x^{2} + y^{2} = C x^{2}$

Show that the equation is homogeneous and solve:

{x cos (\frac{y}{x}) + y sin (\frac{y}{x})} y d x = {y sin (\frac{y}{x}) - x cos (\frac{y}{x})} x d y

Solution

$\frac{d y}{d x} = \frac{v ( cos v + v sin v )}{v sin v - cos v}$ . With $y = v x$ :

$x \frac{d v}{d x} = \frac{v cos v + v ^{2} sin v - v ^{2} sin v + v cos v}{v sin v - cos v} = \frac{2 v cos v}{v sin v - cos v}$ .

$(tan v - \frac{1}{v}) d v = \frac{2 d x}{x} \Rightarrow - lo g ∣ cos v ∣ - lo g ∣ v ∣ = 2 lo g ∣ x ∣ + C \Rightarrow x^{2} v cos v = C^{'}$ .

$x y cos (\frac{y}{x}) = C$

Show that the equation is homogeneous and solve:

x \frac{d y}{d x} - y + x sin (\frac{y}{x}) = 0

Solution

$\frac{d y}{d x} = v - sin v$ ; $y = v x$ : $x \frac{d v}{d x} = - sin v$ .

$cosec v d v = - \frac{d x}{x} \Rightarrow lo g ∣ cosec v - cot v ∣ = - lo g ∣ x ∣ + lo g C \Rightarrow x \cdot \frac{1 - cos v}{sin v} = C$ .

$x [1 - cos (\frac{y}{x})] = C sin (\frac{y}{x})$

Show that the equation is homogeneous and solve:

y d x + x lo g (\frac{y}{x}) d y - 2 x d y = 0

Solution

$\frac{d y}{d x} = \frac{y}{2 x - x lo g \frac{y}{x}} = \frac{v}{2 - lo g v}$ . With $y = v x$ : $x \frac{d v}{d x} = \frac{v ( lo g v - 1 )}{2 - lo g v}$ .

$\frac{2 - lo g v}{v ( lo g v - 1 )} d v = \frac{d x}{x}$ . Put $t = lo g v - 1$ , $d t = \frac{d v}{v}$ : $\frac{1 - t}{t} d t = \frac{d x}{x}$ .

$lo g ∣ t ∣ - t = lo g ∣ x ∣ + C \Rightarrow \frac{t}{x} = K e^{t}$ , and $e^{t} = \frac{y}{e x}$ : $lo g \frac{y}{x} - 1 = \frac{K}{e} y$ .

$cy = lo g \frac{y}{x} - 1$

Show that the equation is homogeneous and solve:

(1 + e^{x / y}) d x + e^{x / y} (1 - \frac{x}{y}) d y = 0

Solution

$\frac{d x}{d y} = - \frac{e ^{x / y} ( 1 - \frac{x}{y} )}{1 + e ^{x / y}}$ . Put $x = v y$ : $v + y \frac{d v}{d y} = - \frac{e ^{v} ( 1 - v )}{1 + e ^{v}}$ .

$y \frac{d v}{d y} = \frac{- e ^{v} + v e ^{v} - v - v e ^{v}}{1 + e ^{v}} = - \frac{v + e ^{v}}{1 + e ^{v}}$ .

$\frac{1 + e ^{v}}{v + e ^{v}} d v = - \frac{d y}{y} \Rightarrow lo g ∣ v + e^{v} ∣ = - lo g ∣ y ∣ + lo g C \Rightarrow y (v + e^{v}) = C$ .

$x + y e^{x / y} = C$

Find the particular solution: (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

Solution

$\frac{d y}{d x} = \frac{y - x}{x + y}$ ; $y = v x$ : $x \frac{d v}{d x} = \frac{v - 1}{1 + v} - v = - \frac{1 + v ^{2}}{1 + v}$ .

$\frac{1 + v}{1 + v ^{2}} d v = - \frac{d x}{x} \Rightarrow tan^{- 1} v + \frac{1}{2} lo g (1 + v^{2}) = - lo g ∣ x ∣ + C$ , i.e. $tan^{- 1} \frac{y}{x} + \frac{1}{2} lo g (x^{2} + y^{2}) = C$ .

At (1, 1): $C = \frac{π}{4} + \frac{1}{2} lo g 2$ .

$lo g (x^{2} + y^{2}) + 2 tan^{- 1} (\frac{y}{x}) = \frac{π}{2} + lo g 2$

Find the particular solution: x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Solution

$\frac{d y}{d x} = - (v + v^{2})$ ; $x \frac{d v}{d x} = - (v^{2} + 2 v)$ .

$\frac{d v}{v ( v + 2 )} = - \frac{d x}{x} \Rightarrow \frac{1}{2} lo g \frac{v}{v + 2} = - lo g ∣ x ∣ + C \Rightarrow \frac{x ^{2} v}{v + 2} = K$ , i.e. $\frac{x ^{2} y}{y + 2 x} = K$ .

At (1, 1): $K = \frac{1}{3}$ .

$y + 2 x = 3 x^{2} y$

Find the particular solution:

[x sin^{2} (\frac{y}{x}) - y] d x + x d y = 0

; y = π/4 when x = 1

Solution

$\frac{d y}{d x} = v - sin^{2} v$ ; $x \frac{d v}{d x} = - sin^{2} v$ .

$cosec^{2} v d v = - \frac{d x}{x} \Rightarrow - cot v = - lo g ∣ x ∣ - C \Rightarrow cot v = lo g ∣ x ∣ + C$ .

At $x = 1$ , $v = \frac{π}{4}$ : $1 = C$ . And $lo g ∣ x ∣ + 1 = lo g ∣ e x ∣$ .

$cot (\frac{y}{x}) = lo g ∣ e x ∣$

Find the particular solution:

\frac{d y}{d x} - \frac{y}{x} + cosec (\frac{y}{x}) = 0

; y = 0 when x = 1

Solution

$y = v x$ : $x \frac{d v}{d x} = - cosec v \Rightarrow sin v d v = - \frac{d x}{x} \Rightarrow cos v = lo g ∣ x ∣ + C$ .

At (1, 0): $C = 1$ .

$cos (\frac{y}{x}) = lo g ∣ e x ∣$

Find the particular solution:

2 x y + y^{2} - 2 x^{2} \frac{d y}{d x} = 0

; y = 2 when x = 1

Solution

$\frac{d y}{d x} = v + \frac{v ^{2}}{2}$ ; $x \frac{d v}{d x} = \frac{v ^{2}}{2}$ .

$\frac{2 d v}{v ^{2}} = \frac{d x}{x} \Rightarrow - \frac{2}{v} = lo g ∣ x ∣ + C \Rightarrow - \frac{2 x}{y} = lo g ∣ x ∣ + C$ . At (1, 2): $C = - 1$ .

$y = \frac{2 x}{1 - lo g ∣ x ∣}$ (x ≠ 0, x ≠ e)

A homogeneous differential equation of the form

\frac{d x}{d y} = h (\frac{x}{y})

can be solved by making the substitution (A) y = vx (B) v = yx (C) x = vy (D) x = v

Solution

(C)

Which of the following is a homogeneous differential equation? (A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (B) (xy) dx – (x³ + y³) dy = 0 (C) (x³ + 2y²) dx + 2xy dy = 0 (D) y² dx + (x² – xy – y²) dy = 0

Solution

In (D) every term has degree 2; (A) has constants, (B) and (C) mix degrees.

(D)

Exercise 9.5

Find the general solution:

\frac{d y}{d x} + 2 y = sin x

Solution

$P = 2$ , IF $= e^{2 x}$ . $y e^{2 x} = \int e^{2 x} sin x d x = \frac{e ^{2 x} ( 2 sin x - cos x )}{5} + C$ .

$y = \frac{1}{5} (2 sin x - cos x) + C e^{- 2 x}$

Find the general solution:

\frac{d y}{d x} + 3 y = e^{- 2 x}

Solution

IF $= e^{3 x}$ : $y e^{3 x} = \int e^{x} d x = e^{x} + C$ .

$y = e^{- 2 x} + C e^{- 3 x}$

Find the general solution:

\frac{d y}{d x} + \frac{y}{x} = x^{2}

Solution

IF $= e^{l o g x} = x$ : $x y = \int x^{3} d x$ .

$x y = \frac{x ^{4}}{4} + C$

Find the general solution:

\frac{d y}{d x} + (sec x) y = tan x

(0 ≤ x < π/2)

Solution

IF $= e^{\int s e c x d x} = sec x + tan x$ .

$y (sec x + tan x) = \int (sec x tan x + tan^{2} x) d x = \int (sec x tan x + sec^{2} x - 1) d x$

$y (sec x + tan x) = sec x + tan x - x + C$

Find the general solution:

cos^{2} x \frac{d y}{d x} + y = tan x

(0 ≤ x < π/2)

Solution

$\frac{d y}{d x} + sec^{2} x \cdot y = tan x sec^{2} x$ ; IF $= e^{t a n x}$ .

$y e^{t a n x} = \int tan x sec^{2} x e^{t a n x} d x = \int t e^{t} d t$ (t = tan x) $= (t - 1) e^{t} + C$ .

$y = (tan x - 1) + C e^{- t a n x}$

Find the general solution:

x \frac{d y}{d x} + 2 y = x^{2} lo g x

Solution

$\frac{d y}{d x} + \frac{2}{x} y = x lo g x$ ; IF $= x^{2}$ .

$x^{2} y = \int x^{3} lo g x d x = \frac{x ^{4}}{4} lo g x - \frac{x ^{4}}{16} + C$ .

$y = \frac{x ^{2}}{16} (4 lo g x - 1) + C x^{- 2}$

Find the general solution:

x lo g x \frac{d y}{d x} + y = \frac{2}{x} lo g x

Solution

$\frac{d y}{d x} + \frac{y}{x lo g x} = \frac{2}{x ^{2}}$ ; IF $= e^{\int \frac{d x}{x l o g x}} = lo g x$ .

$y lo g x = \int \frac{2 lo g x}{x ^{2}} d x = 2 [- \frac{lo g x}{x} + \int \frac{d x}{x ^{2}}] = - \frac{2}{x} (1 + lo g x) + C$ .

$y lo g x = - \frac{2}{x} (1 + lo g x) + C$

Find the general solution: (1 + x²) dy + 2xy dx = cot x dx (x ≠ 0)

Solution

$\frac{d y}{d x} + \frac{2 x}{1 + x ^{2}} y = \frac{cot x}{1 + x ^{2}}$ ; IF $= 1 + x^{2}$ .

$y (1 + x^{2}) = \int cot x d x$ .

$y (1 + x^{2}) = lo g ∣ sin x ∣ + C$

Find the general solution:

x \frac{d y}{d x} + y - x + x y cot x = 0

(x ≠ 0)

Solution

$\frac{d y}{d x} + (\frac{1}{x} + cot x) y = 1$ ; IF $= e^{l o g x + l o g s i n x} = x sin x$ .

$x y sin x = \int x sin x d x = - x cos x + sin x + C$ .

$y = \frac{1}{x} - cot x + \frac{C}{x sin x}$

Find the general solution:

(x + y) \frac{d y}{d x} = 1

Solution

$\frac{d x}{d y} = x + y \Rightarrow \frac{d x}{d y} - x = y$ (linear in x); IF $= e^{- y}$ .

$x e^{- y} = \int y e^{- y} d y = - y e^{- y} - e^{- y} + C \Rightarrow x = - y - 1 + C e^{y}$ .

$x + y + 1 = C e^{y}$

Find the general solution: y dx + (x – y²) dy = 0

Solution

$\frac{d x}{d y} + \frac{x}{y} = y$ ; IF $= y$ . $x y = \int y^{2} d y = \frac{y ^{3}}{3} + C$ .

$x = \frac{y ^{2}}{3} + \frac{C}{y}$

Find the general solution:

(x + 3 y^{2}) \frac{d y}{d x} = y

(y > 0)

Solution

$\frac{d x}{d y} = \frac{x + 3 y ^{2}}{y} \Rightarrow \frac{d x}{d y} - \frac{x}{y} = 3 y$ ; IF $= \frac{1}{y}$ . $\frac{x}{y} = \int 3 d y = 3 y + C$ .

$x = 3 y^{2} + C y$

Find a particular solution:

\frac{d y}{d x} + 2 y tan x = sin x

; y = 0 when x = π/3

Solution

IF $= e^{2 l o g s e c x} = sec^{2} x$ . $y sec^{2} x = \int sin x sec^{2} x d x = \int sec x tan x d x = sec x + C$ .

At $x = \frac{π}{3}$ : $0 = 2 + C \Rightarrow C = - 2$ . So $y sec^{2} x = sec x - 2$ .

$y = cos x - 2 cos^{2} x$

Find a particular solution:

(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x ^{2}}

; y = 0 when x = 1

Solution

$\frac{d y}{d x} + \frac{2 x}{1 + x ^{2}} y = \frac{1}{( 1 + x ^{2} ) ^{2}}$ ; IF $= 1 + x^{2}$ . $y (1 + x^{2}) = tan^{- 1} x + C$ .

At $x = 1$ : $0 = \frac{π}{4} + C$ .

$y (1 + x^{2}) = tan^{- 1} x - \frac{π}{4}$

Find a particular solution:

\frac{d y}{d x} - 3 y cot x = sin 2 x

; y = 2 when x = π/2

Solution

IF $= e^{- 3 l o g s i n x} = \frac{1}{sin ^{3} x}$ . $\frac{y}{sin ^{3} x} = \int \frac{2 sin x cos x}{sin ^{3} x} d x = \int \frac{2 cos x}{sin ^{2} x} d x = - \frac{2}{sin x} + C$ .

At $x = \frac{π}{2}$ : $2 = - 2 + C \Rightarrow C = 4$ .

$y = 4 sin^{3} x - 2 sin^{2} x$

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Solution

$\frac{d y}{d x} = x + y \Rightarrow \frac{d y}{d x} - y = x$ ; IF $= e^{- x}$ .

$y e^{- x} = \int x e^{- x} d x = - (x + 1) e^{- x} + C$ . Through (0, 0): $C = 1$ . So $y = - (x + 1) + e^{x}$ .

$x + y + 1 = e^{x}$

Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Solution

$x + y - \frac{d y}{d x} = 5 \Rightarrow \frac{d y}{d x} - y = x - 5$ ; IF $= e^{- x}$ .

$y e^{- x} = \int (x - 5) e^{- x} d x = - (x - 5) e^{- x} - e^{- x} + C = (4 - x) e^{- x} + C$ , so $y = 4 - x + C e^{x}$ .

At (0, 2): $2 = 4 + C \Rightarrow C = - 2$ .

$y = 4 - x - 2 e^{x}$

The Integrating Factor of the differential equation

x \frac{d y}{d x} - y = 2 x^{2}

is (A) e⁻ˣ (B) e⁻ʸ (C) 1/x (D) x

Solution

$\frac{d y}{d x} - \frac{y}{x} = 2 x$ : $P = - \frac{1}{x}$ , IF $= e^{- l o g x} = \frac{1}{x}$ .

(C)

The Integrating Factor of the differential equation

(1 - y^{2}) \frac{d x}{d y} + y x = a y

(–1 < y < 1) is (A)

\frac{1}{y ^{2} - 1}

(B)

\frac{1}{y ^{2} - 1}

(C)

\frac{1}{1 - y ^{2}}

(D)

\frac{1}{1 - y ^{2}}

Solution

$\frac{d x}{d y} + \frac{y}{1 - y ^{2}} x = \frac{a y}{1 - y ^{2}}$ : IF $= e^{\int \frac{y}{1 - y ^{2}} d y} = e^{- \frac{1}{2} l o g (1 - y^{2})} = \frac{1}{1 - y ^{2}}$ .

(D)

Miscellaneous Exercise on Chapter 9

Indicate order and degree (if defined): (i)

\frac{d ^{2} y}{d x ^{2}} + 5 x (\frac{d y}{d x})^{2} - 6 y = lo g x

(ii)

(\frac{d y}{d x})^{3} - 4 (\frac{d y}{d x})^{2} + 7 y = sin x

(iii)

\frac{d ^{4} y}{d x ^{4}} - sin (\frac{d ^{3} y}{d x ^{3}}) = 0

Solution

(i) Highest derivative $y^{''}$ , power 1. (ii) Only $y^{'}$ appears; highest power 3. (iii) $sin (y^{'''})$ is not polynomial.

(i) Order 2; degree 1 (ii) Order 1; degree 3 (iii) Order 4; degree not defined

Verify that the given function is a solution of the corresponding differential equation: (i) xy = aeˣ + be⁻ˣ + x² :

x \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0

(ii) y = eˣ (a cos x + b sin x) :

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 2 y = 0

(iii) y = x sin 3x :

\frac{d ^{2} y}{d x ^{2}} + 9 y - 6 cos 3 x = 0

(iv) x² = 2y² log y :

(x^{2} + y^{2}) \frac{d y}{d x} - x y = 0

Solution

(i) Differentiate: $y + x y^{'} = a e^{x} - b e^{- x} + 2 x$ . Again: $2 y^{'} + x y^{''} = a e^{x} + b e^{- x} + 2 = (x y - x^{2}) + 2$ . So $x y^{''} + 2 y^{'} - x y + x^{2} - 2 = 0$ .

(ii) $y^{'} = e^{x} [(a + b) cos x + (b - a) sin x]$ , $y^{''} = e^{x} [2 b cos x - 2 a sin x]$ .

$y^{''} - 2 y^{'} + 2 y = e^{x} [(2 b - 2 a - 2 b + 2 a) cos x + (- 2 a - 2 b + 2 a + 2 b) sin x] = 0$ .

(iii) $y^{'} = sin 3 x + 3 x cos 3 x$ , $y^{''} = 6 cos 3 x - 9 x sin 3 x = 6 cos 3 x - 9 y$ . So $y^{''} + 9 y - 6 cos 3 x = 0$ .

(iv) Differentiate: $2 x = y^{'} (4 y lo g y + 2 y) = 2 y y^{'} (2 lo g y + 1)$ . From the equation, $2 lo g y = \frac{x ^{2}}{y ^{2}}$ , so $2 x = 2 y y^{'} \cdot \frac{x ^{2} + y ^{2}}{y ^{2}}$ , giving $y^{'} = \frac{x y}{x ^{2} + y ^{2}}$ .

All four verified.

Prove that x² – y² = c (x² + y²)² is the general solution of differential equation (x³ – 3xy²) dx = (y³ – 3x²y) dy, where c is a parameter.

Solution

Differentiate: $2 x - 2 y y^{'} = 2 c (x^{2} + y^{2}) (2 x + 2 y y^{'}) \Rightarrow x - y y^{'} = 2 c (x^{2} + y^{2}) (x + y y^{'})$ .

Eliminate c using $c = \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ : $(x - y y^{'}) (x^{2} + y^{2}) = 2 (x^{2} - y^{2}) (x + y y^{'})$ .

$x^{3} + x y^{2} - y y^{'} (x^{2} + y^{2}) = 2 x^{3} - 2 x y^{2} + 2 y y^{'} (x^{2} - y^{2})$

$y^{'} (- x^{2} y - y^{3} - 2 x^{2} y + 2 y^{3}) = x^{3} - 3 x y^{2} \Rightarrow y^{'} (y^{3} - 3 x^{2} y) = x^{3} - 3 x y^{2}$ .

This is $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ . Proved.

Find the general solution of the differential equation

\frac{d y}{d x} + \frac{1 - y ^{2}}{1 - x ^{2}} = 0

Solution

$\frac{d y}{1 - y ^{2}} = - \frac{d x}{1 - x ^{2}} \Rightarrow sin^{- 1} y = - sin^{- 1} x + C$ .

$sin^{- 1} y + sin^{- 1} x = C$

Show that the general solution of the differential equation

\frac{d y}{d x} + \frac{y ^{2} + y + 1}{x ^{2} + x + 1} = 0

is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.

Solution

$\frac{d y}{y ^{2} + y + 1} + \frac{d x}{x ^{2} + x + 1} = 0$ . Since $\int \frac{d x}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} = \frac{2}{3} tan^{- 1} \frac{2 x + 1}{3}$ :

$tan^{- 1} \frac{2 y + 1}{3} + tan^{- 1} \frac{2 x + 1}{3} = C$ .

Take tan of both sides:

$\frac{\frac{2 x + 2 y + 2}{3}}{1 - \frac{( 2 x + 1 ) ( 2 y + 1 )}{3}} = \frac{2 3 ( x + y + 1 )}{2 - 2 x - 2 y - 4 x y} = \frac{3 ( x + y + 1 )}{1 - x - y - 2 x y} = tan C$

$x + y + 1 = A (1 - x - y - 2 x y)$ with $A = \frac{tan C}{3}$ . Shown.

Find the equation of the curve passing through the point (0, π/4) whose differential equation is sin x cos y dx + cos x sin y dy = 0.

Solution

Divide by $cos x cos y$ : $tan x d x + tan y d y = 0 \Rightarrow lo g ∣ sec x ∣ + lo g ∣ sec y ∣ = lo g C \Rightarrow sec x sec y = C$ .

At $(0, \frac{π}{4})$ : $C = 2$ .

$cos y = \frac{sec x}{2}$

Find the particular solution of the differential equation (1 + e²ˣ) dy + (1 + y²) eˣ dx = 0, given that y = 1 when x = 0.

Solution

$\frac{d y}{1 + y ^{2}} = - \frac{e ^{x} d x}{1 + e ^{2 x}} \Rightarrow tan^{- 1} y = - tan^{- 1} e^{x} + C$ .

At $x = 0$ , $y = 1$ : $\frac{π}{4} = - \frac{π}{4} + C \Rightarrow C = \frac{π}{2}$ .

$tan^{- 1} y + tan^{- 1} (e^{x}) = \frac{π}{2}$

Solve the differential equation

y e^{x / y} d x = (x e^{x / y} + y^{2}) d y

(y ≠ 0).

Solution

$\frac{d x}{d y} = \frac{x}{y} + y e^{- x / y}$ . Put $x = v y$ : $v + y \frac{d v}{d y} = v + y e^{- v} \Rightarrow e^{v} d v = d y \Rightarrow e^{v} = y + C$ .

$e^{x / y} = y + C$

Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

Solution

$t = x - y$ , $d t = d x - d y$ , so $d x + d y = 2 d x - d t$ : $t (2 d x - d t) = d t \Rightarrow 2 d x = (1 + \frac{1}{t}) d t$ .

$2 x = t + lo g ∣ t ∣ + C \Rightarrow 2 x = x - y + lo g ∣ x - y ∣ + C \Rightarrow x + y = lo g ∣ x - y ∣ + C$ .

At (0, −1): $- 1 = 0 + C$ .

$lo g ∣ x - y ∣ = x + y + 1$

Solve the differential equation

[\frac{e ^{- 2 x}}{x} - \frac{y}{x}] \frac{d x}{d y} = 1

(x ≠ 0).

Solution

$\frac{d y}{d x} + \frac{y}{x} = \frac{e ^{- 2 x}}{x}$ ; IF $= e^{\int x^{- 1/2} d x} = e^{2 x}$ .

$y e^{2 x} = \int \frac{d x}{x}$ .

$y e^{2 x} = 2 x + C$

Find a particular solution of the differential equation

\frac{d y}{d x} + y cot x = 4 x cosec x

(x ≠ 0), given that y = 0 when x = π/2.

Solution

IF $= e^{\int c o t x d x} = sin x$ . $y sin x = \int 4 x d x = 2 x^{2} + C$ .

At $x = \frac{π}{2}$ : $0 = \frac{π ^{2}}{2} + C$ .

$y sin x = 2 x^{2} - \frac{π ^{2}}{2}$ (sin x ≠ 0)

Find a particular solution of the differential equation

(x + 1) \frac{d y}{d x} = 2 e^{- y} - 1

, given that y = 0 when x = 0.

Solution

$\frac{d y}{2 e ^{- y} - 1} = \frac{d x}{x + 1} \Rightarrow \frac{e ^{y} d y}{2 - e ^{y}} = \frac{d x}{x + 1} \Rightarrow - lo g ∣2 - e^{y} ∣ = lo g ∣ x + 1∣ + C \Rightarrow (2 - e^{y}) (x + 1) = K$ .

At (0, 0): $K = 1$ . So $e^{y} = 2 - \frac{1}{x + 1} = \frac{2 x + 1}{x + 1}$ .

$y = lo g \frac{2 x + 1}{x + 1}$ , x ≠ −1

The general solution of the differential equation

\frac{y d x - x d y}{y} = 0

is (A) xy = C (B) x = Cy² (C) y = Cx (D) y = Cx²

Solution

$y d x = x d y \Rightarrow \frac{d x}{x} = \frac{d y}{y} \Rightarrow lo g ∣ y ∣ = lo g ∣ x ∣ + lo g C$ .

(C)

The general solution of a differential equation of the type

\frac{d x}{d y} + P_{1} x = Q_{1}

is (A)

y e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C

(B)

y e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C

(C)

x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C

(D)

x e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C

Solution

The equation is linear in x with y as the independent variable, so IF $= e^{\int P_{1} d y}$ and x is multiplied by it.

(C)

The general solution of the differential equation eˣ dy + (y eˣ + 2x) dx = 0 is (A) x eʸ + x² = C (B) x eʸ + y² = C (C) y eˣ + x² = C (D) y eʸ + x² = C

Solution

$e^{x} d y + y e^{x} d x = d (y e^{x})$ , so $d (y e^{x}) + 2 x d x = 0 \Rightarrow y e^{x} + x^{2} = C$ .

(C)

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