Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.9 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.9

Choose the most suitable answer from the given four alternatives:

Question 1.

The order and degree of the differential equation \(\frac { d^2y }{ dx^2 }\) + (\(\frac { dy }{ dx }\))^{1/3} + x^{1/4} = o are respectively

(a) 2, 3

(b) 3, 3

(c) 2, 6

(d) 2, 4

Solution:

(a) 2, 3

Hint:

In this equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) & its power is 3.

∴ Its order = 2 & degree = 3

Question 2.

The differential equation representing the family of curves y = A cos (x + B), where A and B are parameters, is

(a) \(\frac { d^2y }{ dx^2 }\) – y = 0

(b) \(\frac { d^2y }{ dx^2 }\) + y = 0

(c) \(\frac { d^2y }{ dx^2 }\) = 0

(d) \(\frac { d^2x }{ dy^2 }\) = 0

Solution:

(b) \(\frac { d^2y }{ dx^2 }\) + y = 0

Hint:

Given equation is y = A cos (x + B) …….. (1)

where A & B are parameters.

Differentiating equation (1) twice successively, because we have two arbitrary constants.

\(\frac { dy }{ dx }\) = A sin (x + B)

Again differentiating \(\frac { d^2y }{ dx^2 }\) = -A cos(x + B) = -y

∵ y = A cos (x + B

∵ \(\frac { d^2y }{ dx^2 }\) + y = 0 as the required differential equation.

Question 3.

The order and degree of the differential equation, \(\sqrt { sin x }\) (dx + dy) = \(\sqrt { cos x }\) (dx – dy) is

(a) 1, 2

(b) 2, 2

(c) 1, 1

(d) 2, 1

Solution:

(c) 1, 1

Hint:

Given \(\sqrt { sin x }\) (dx + dy) = \(\sqrt { cos x }\) (dx – dy)

divide by dx on both sides, we get

In this equation, the highest, order derivative is \(\frac { dy }{ dx }\) & its power is 1.

∴ Its order = 1 & degree = 1

Question 4.

The order of the differential equation of all circles with centre at (h, k) and radius ‘a’ is

(a) 2

(b) 3

(c) 4

(d) 1

Solution:

(a) 2

Hint:

We know that equation of a circle be (x – h)² + (y – k)² = a².

Here we have two constants. Therefore, Its order is 2.

Question 5.

The differential equation of the family of curves y = Ae^{x} + Be^{-x}, where A and B are arbitrary constants is

(a) \(\frac { d^2y }{ dx^2 }\) + y = 0

(b) \(\frac { d^2y }{ dx^2 }\) – y = 0

(c) \(\frac { dy }{ dx }\) + y = 0

(d) \(\frac { dy }{ dx }\) – y = 0

Solution:

(b) \(\frac { d^2y }{ dx^2 }\) – y = 0

Hint:

Given y = Ae^{x} + Be^{-x} ……… (1)

where A & B are arbitrary constants.

Differentiate equation(1) twice continuously, we get

(∵ Two constants so differentiate twice)

\(\frac { d^2y }{ dx^2 }\) – y = 0 as the required differential equation.

Question 6.

The general solution pf the differential equation \(\frac { dy }{ dx }\) = \(\frac { y }{ x }\) is

(a) xy = k

(b) y = k log x

(c) y = kx

(d) log y = kr

Solution:

(c) y = kx

Hint:

Given \(\frac { dy }{ dx }\) = \(\frac { y }{ x }\)

The equation can be written as

\(\frac { dy }{ y }\) = \(\frac { dx }{ x }\)

Integrating on both sides, we get

∫\(\frac { dy }{ y }\) = ∫\(\frac { dy }{ x }\)

log y = log x + log k

log y = log k x [∵ log m + log n = log mn]

Remove log, we get

y = kx is a required differential equation

Question 7.

The solution of the differential equation 2x\(\frac { dy }{ dx }\) – y = 3 represents

(a) straight lines

(b) circles

(c) parabola

(d) ellipse

Solution:

(c) parabola

Hint:

2 log (3 + y) = log x + log k

log (3 + y)² = log kx

Remove log, we get

(3 + y)² = kx is a solution of the differential equation which is a Parabola.

Question 8.

The solution of \(\frac { dy }{ dx }\) + p(x) y = 0 is

(a) y = ce^{∫pdx}

(b) y = ce^{-∫pdx}

(c) x = ce^{-∫pdy}

(d) x = ce^{∫pdy}

Solution:

(b) y = ce^{-∫pdx}

Hint:

Given \(\frac { dy }{ dx }\) + p(x) y = 0

\(\frac { dy }{ dx }\) = -p(x)y

The equation can be written as

\(\frac { dy }{ y }\) = -p(x) dx

Taking integration on both sides, we get

∴ y = ce^{-∫pdx} is a solution of the given differential equation.

Question 9.

The integrating factor of the differential equation \(\frac { dy }{ dx }\) + y = \(\frac { 1+y }{ x }\) is

(a) \(\frac { x }{ e^x }\) + y = 0

(b) \(\frac { e^x }{ x }\) – y = 0

(c) λe^{x}

(d) e^{x}

Solution:

(b) \(\frac { e^x }{ x }\) – y = 0

Hint:

Given differential equation is

Question 10.

The Integrating factor of the differential equation \(\frac { dy }{ dx }\) + p(x) y = Q(x) is x, then p(x)

(a) x

(b) \(\frac { x^2 }{ 2 }\)

(c) \(\frac { 1 }{ x }\)

(d) \(\frac { 1 }{ x^2 }\)

Solution:

(c) \(\frac { 1 }{ x }\)

Hint:

The given differential equation is

\(\frac { dy }{ dx }\) + p(x) y = Q(x)

Integrating factor e^{∫pdx} = x

Taking log on both sides, we get ∫p dx = log x

Now differentiating, we get I

\(\frac { d }{ dx }\) [∫p dx]= \(\frac { d }{ dx }\) [log x]

∴ p = \(\frac { 1 }{ x }\)

Question 11.

The degree of the differential equation

y(x) = 1 + \(\frac { dy }{ dx }\) + \(\frac { 1 }{ 1.2 }\) (\(\frac { dy }{ dx }\))² + \(\frac { 1 }{ 1.2.3 }\) (\(\frac { dy }{ dx }\))³ + ……. is

(a) 2

(b) 3

(c) 1

(d) 4

Solution:

(c) 1

Hint:

Given

y(x) = 1 + \(\frac { dy }{ dx }\) + \(\frac { 1 }{ 1.2 }\) (\(\frac { dy }{ dx }\))² + \(\frac { 1 }{ 1.2.3 }\) (\(\frac { dy }{ dx }\))³ + ……

Here the highest derivative term is \(\frac { dy }{ dx }\) and its power is 1.

∴ Degree of the differential equation is 1.

Question 12.

If p and q are the order and degree of the differential equation y \(\frac { dy }{ dx }\) +x³ (\(\frac { d^2y }{ dx^2 }\)) + xy = cos x, when

(a) p < q

(b) p = q

(c) p > q

(d) p exists and q does not exist

Solution:

(c) p > q

Hint:

Given equation is y\(\frac { dy }{ dx }\) +x³ (\(\frac { d^2y }{ dx^2 }\)) + xy = cos x

The highest order derivative of the differential equation is \(\frac { d^2y }{ dx^2 }\) and its degree is 1.

∴ Its order is p = 2 and degree q = 1

∴ p > q

Question 13.

The solution of the differential equation \(\frac { dy }{ dx }\) + \(\frac { 1 }{ \sqrt{1-x^2} }\) = 0 is

(a) y + sin^{-1} x = c

(b) x + sin^{-1} y = 0

(c) y² + 2sin^{-1} x = c

(d) x² + 2sin^{-1} y = 0

Solution:

(a) y + sin^{-1} x = c

Hint:

The given equation is \(\frac { dy }{ dx }\) = –\(\frac { 1 }{ \sqrt{1-x^2} }\)

The equation can be written as

dy = \(\frac { 1 }{ \sqrt{1-x^2} }\)

Integrating on both sides, we get

∫dy = -∫\(\frac { dx }{ \sqrt{1-x^2} }\)

y = -sin^{-1} (x) + C

y + sin^{-1} (x) = C

∴ y + sin^{-1} (x) = C is a solution of the given differential equation.

Question 14.

The solution of the differential equation \(\frac { dy }{ dx }\) = 2xy is

(a) y = Ce^{x²}

(b) y = 2x² + C

(c) y = Ce^{-x²} + C

(d) y = x² + C

Solution:

(a) y = Ce^{x²}

Hint:

Given

\(\frac { dy }{ dx }\) = 2xy

The equation can be written as

\(\frac { dy }{ y }\) = 2xdx

Taking integration on both sides, we get

∴ y = Ce^{x²} is a solution to the differential equation.

Question 15.

The general solution of the differential equation log(\(\frac { dy }{ dx }\)) = x + y is

(a) e^{x} + e^{y} = C

(b) e^{x} + e^{-y} = C

(c) e^{-x} + e^{y} = C

(d) e^{-x} + e^{-y} = C

Solution:

(b) e^{x} + e^{-y} = C

Hint:

Given differential equation is log \(\frac { dy }{ dx }\) = x + y,

log \(\frac { dy }{ dx }\) = x + y

\(\frac { dy }{ dx }\) = e^{x+y} = e^{x} e^{y}

\(\frac { dy }{ dx }\) = e^{x} e^{y}

The given equation can be written as

\(\frac { dy }{ dx }\) = e^{x} e^{y}

e^{-y} dy = e^{x} dx

Taking integration on both sides, we get

∫ e^{-y} dy = ∫ e^{x} dx

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

-e^{-y} = e^{x} + C

-e^{x} – e^{-y} = C

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

e^{x} + e^{-y} = -C

∴ e^{x} + e^{-y} = C Where – C = C

∴ e^{x} + e^{-y} = C is a solution of the given differential equation.

Question 16.

The solution of \(\frac { dy }{ dx }\) = 2^{y-x} is

(a) 2^{x} + 2^{y} = C

(b) 2^{x} – 2^{y} = C

(c) \(\frac { 1 }{ 2^x }\) – \(\frac { 1 }{ 2^y }\) = C

(d) x + y = C

Solution:

(c) \(\frac { 1 }{ 2^x }\) – \(\frac { 1 }{ 2^y }\) = C

Hint:

Given

\(\frac { dy }{ dx }\) = 2^{y-x}

The equation can be written as

\(\frac { 1 }{ 2^x }\) – \(\frac { 1 }{ 2^y }\) = C is a solution of the given differential equation.

Question 17.

The solution of the differential equation

\(\frac { dy }{ dx }\) = \(\frac { y }{ x }\) + \(\frac { ∅(\frac { y }{ x }) }{ ∅(\frac { y }{ x }) }\) is

(a) x∅(\(\frac { y }{ x }\)) = k

(b) ∅(\(\frac { y }{ x }\)) = kx

(c) y∅(\(\frac { y }{ x }\)) = k

(d) ∅(\(\frac { y }{ x }\)) = ky

Solution:

(b) ∅\(\frac { y }{ x }\) = kx

Hint:

log ∅(v) = log x + log k

log ∅(v) = log xk

∅(v) = kx

∅(y/x) = kx

Question 18.

If sin x is the integrating factor of the linear differential equation \(\frac { dy }{ dx }\) + Py = Q, then P is

(a) log sin x

(b) cos x

(c) tan x

(d) cot x

Solution:

(d) cot x

Hint:

Given integrating factor

e^{∫pdx} = sin x

∫pdx = log sin x ……. (1)

Differential equation (1) with respect to x, we get

∴ The value of P is cot x

Question 19.

The number of arbitrary constants in the general solutions of order n and n + 1 is respectively.

(a) n – 1, n

(b) n, n + 1

(c) n + 1, n + 2

(d) n + 1, n

Solution:

(b) n, n + 1

Hint:

If one arbitrary constant, Differentiate one time, so order is 1.

If two arbitrary constants, Differentiate two times, so order is 2.

.

.

.

If n arbitrary constants, Differentiate n times, so the order is n.

If n + 1 arbitrary constants, Differentiate n + 1 times, so order is n + 1.

Question 20.

The number of arbitrary constants in the particular solution of a differential equation of third order is

(a) 3

(b) 2

(c) 1

(d) 0

Solution:

(d) 0

Question 21.

Integrating factor of the differential equation

\(\frac { dy }{ dx }\) = \(\frac { x+y+1 }{ x+1 }\) is

(a) \(\frac { 1 }{ x+1 }\)

(b) x+ 1

(c) \(\frac { 1 }{ \sqrt{x+1} }\)

(d) \(\sqrt { x+1 }\)

Solution:

(a) \(\frac { 1 }{ x+1 }\)

Hint:

Question 22.

The population P in any year t is such that the rate of increase in the population is proportional to the population. Then

(a) P = Ce^{kt}

(b) P = Ce^{-kt}

(c) P = Ckt

(d) Pt = C

Solution:

(a) \(\frac { 1 }{ x+1 }\)

Hint:

Question 23.

P is the amount of certain substance left in after time t. If the rate of evaporation of the substance is proportional to the amount remaining, then

(a) P = Ce^{kt}

(b) P = Ce^{-kt}

(c) P = Ckt

(d) Pt = C

Solution:

(b) P = Ce^{-kt}

Hint:

Question 24.

If the solution of the differential equation \(\frac { dy }{ dx }\) = \(\frac { ax+3 }{ 2y+f }\) represents a circle, then the value of a is

(a) 2

(b) -2

(c) 1

(d) -1

Solution:

(b) -2

Hint:

Given \(\frac { dy }{ dx }\) = \(\frac { ax+3 }{ 2y+f }\)

The equation can be written as

(2y + f) dy = (ax + 3) dx ……. (1)

Integrating equation (1) on both sides, we get

It is solution of the given differential equation.

Since this solution represents a circle,

co-efficient of x² = co-efficient of y²

i.e) \(\frac { a }{ 2 }\) = -1

a = -2

Question 25.

The slope at any point of a curve y = f(x) is given by \(\frac { dy }{ dx }\) = 3x² and it passes through (-1, 1). Then the equation of the curve is

(a) y = x³ + 2

(b) y = 3x² + 4

(c) y = 3x³ + 4

(d) y = x³ + 5

Solution:

(a) y = x³ + 2

Hint:

Given differential equation is \(\frac { dy }{ dx }\) = 3x²

The equation can be written as dy = 3x² dx ……… (1)

Integrating equation (1) on both sides, we get

∫dy = ∫ 3x²dx

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

y = x³ + C …….. (2)

Since it passes through (-1, 1)

So, y = x³ + C becomes

1 =(-1)³ + C

1 = -1 + C

1 + 1 = C

∴ C = 2

Substituting C value in equation (2), We get The equation of the curve is y = x³ + 2