Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.9

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