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These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-12-ex-12-1/

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.1

Ex 12.1 Class 12 Question 1.
Maximize Z = 3x + 4y
subject to the constraints:
x + y ≤ 4, x ≥ 0, y ≥ 0.
Solution:
The objective function is Z = 3x + 4y
The constraints are x + y ≤ 4, x ≥ 0, y ≥ 0.

∴ Maximum value of Z is 16 at B (0, 4).

Question 2.
Minimize Z = – 3x + 4y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Solution:
The objective function is Z = – 3x + 4y The constraints are x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0

∴ Maximum value of Z is – 12 at A (4, 0).

Question 3.
Maximize Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
Solution:
The objective function is Z = 5x + 3y
The constraints are 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
The feasible region is shaded in the figure. The point B is obtained by solving the equation 5x + 2y = 10 and 3x + 5y = 15.

Question 4.
Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Solution:
The objective function is Z = 3x + 5y
The constraints are x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
The feasible region is shaded in the figure.

From the table the minimum value of Z is 7. Since the feasible region is unbounded, 7 may or may not be the minimum value of Z. Consider the graph of the inequality 3x + 5y < 7. This half plane has no point in common with the feasible region. Hence the minimum value of Z is 7 at B (\(\frac { 3 }{ 2 }\), \(\frac { 1 }{ 2 }\))

Question 5.
Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Solution:

The objective function is Z = 3x + 5y
The constraints are x + 2y < 10,
3x + y ≤ 15, x, y ≥ 0.
The feasible region is shaded in the figure. We use the corner point method to find the maximum of Z
Maximum value of Z is 18 at B(4, 3)

Question 6.
Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Solution:
Consider 2x + y ≥ 3
Let 2x + y = 3
⇒ y = 3 – 2x

(0,0) is not contained in the required half plane as (0, 0) does not satisfy the in equation 2x + y ≥ 3.
Again consider x + 2y ≥ 6
Let x + 2y = 6
⇒ \(\frac { x }{ 6 } +\frac { y }{ 3 }\) = 1
Here also (0,0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its comers are A (6,0) and B (0,3). At A, Z = 6 + 0 = 6
At B, Z = 0 + 2 × 3 = 6
We see that at both points the value of Z = 6 which is minimum. In fact at every point on the line AB makes Z = 6 which is also minimum.

Question 7.
Minimise and Maximise Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0
Solution:
The objective function is Z = 5x + 10v.
The constraints are x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0

The feasible region is shaded in the figure. We use the comer point method to find the maximum/minimum value of Z.

Minimum value of Z is 300 at A(60,0).
The maximum value of Z is 600 at all points on the line joining B(120, 0) and C(60, 30).

Question 8.
Minimize and maximize Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
Solution:
The objective function is Z = x + 2y
The constraints are x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0

The feasible region is shaded in the figure. We use corner point method to find maxi-mum/minimum value of Z.

Maximum value of Z is 400 at B(0,200). The minimum value of Z is 100 at all points of the line joining the points (0,50) and (20,40).

Question 9.
Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
Solution:

The objective function is Z = – x + 2y
The constraints are x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
The feasible region is shaded in the figure.

From the table the maximum value of Z is 1. Since the feasible region is unbounded 1 may or may not be the maximum value of Z
Consider the graph of the inequality – x + 2y > 1. This half plane has points common with the feasible region. Hence there is no maximum value for Z.

Question 10.
Maximize Z = x + y subject to x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0
Solution:
The objective function is Z = x + y
The constraints are x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0

There is no point satisfying the constraints simultaneously. Thus the problem has no feasible region. Hence no maximum value for Z.

These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-12-ex-12-2/

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.2

Ex 12.2 Class 12 NCERT Solutions Question 1.
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/ kg of Vitamin A and 5 units/ kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Solution:
Let the mixture contain x units of food P and y units of food Q. We have the data as

Vitamin A constaint: 3x + Ay ≥ 8
Vitamin B constraint: 5x + 2y ≥ 11
Cost function : Z = 60x + 80y
The L.P.P is Minimise Z = 60x + 80y
subject to the constraints 3x + 4y ≥8, 5x + 2y ≥ 11, x, y ≥ 0.

The feasible region is shaded in the figure and is unbounded.
We use comer point method to find the mini-mum of Z
From the table, the minimum value of Z is 160. Since the feasible region is unbounded j 160 may or may not be the minimum value i of Z. Consider the graph of the inequality 60x + 80y ≤ 160
i.e., 3x + 4y ≤ 8
This half plane has no point in common with the feasible region. Then the minimum value of Z is 160 at the points on the line segment joining A\(\frac { 8 }{ 3 }\), 0) and B(2, \(\frac { 1 }{ 2 }\))

Exercise 12.2 Class 12 Maths Ncert Solutions In Hindi Question 2.
One kind of cake requires 200 g of flour and 25g of fat, and another kind of cake requires 100 g of flour and 50 g of fat Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients, used in making the cakes.
Solution:
Let x be the number of cakes of type I and y be the number of cakes of type II.
We make use of the following table to write the L.P.P.
We have the data as

Flour constraint: 200x + 100y ≤ 5000
i.e., 2x + y ≤ 50
Fat constraint: 25x + 50y ≤ 1000
i.e., x + 2y ≤ 40
Total number of cakes : Z = x + y
The L.P.P is Maximise Z = x + y
subject to the constraints
2x + y ≤ 50, x + 2y ≤ 40, x, y >0.

∴ Maximum value of Z is 30 at B(20, 10).
Hence maximum number of cakes is 30, of which 20 are of type 1 and 10 are of type II.

Chapter 12 Exercise 12.2 NCERT Solutions Question 3.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.
i. What number of rackets and bats must be made if the factory is to work at full capacity?
ii. If the profit on a racket and on a bat is ₹ 20 and ₹ 10 respectively, find the maximum profit of the factory when it works at full capacity.
Solution:
i. Let x be the number of tennis rackets and that y be the cricket bats.

Machine constraint: 1.5x + 3y ≤ 42
i.e., x + 2y ≤ 28
Craftman’s constraint: 3x + y ≤ 24
Profit function : Z = 20x + 10y
The L.P.P is Maximise Z = 20x + 10y subject to the constraints
x + 2y ≤ 28, 3x + y ≤ 24, x, y ≥ 0

∴ Maximum value of Z is 200 at B(4, 12).
Hence 4 tennis rackets and 12 cricket bats can be made if the factory is to work at full capacity.

ii. The maximum profit is ₹ 200/-

Ncert Solutions For Class 12 Maths Chapter 12 Question 4.
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ₹ 17.50 per package on nuts and ₹ 7.00 per package on bolts. How many package of each should be produced each day so as to maximise his profit, if he operates his machine for at the most 12 hours a day?
Solution:
Let x and y be the number of packages of nuts and bolts.
∴ We have the data as

Machine A constraint: x + 3y ≤ 12
Machine B constraint: 3x + y ≤ 12
Profit function : Z = 17.5x + 7y
The L.P.P is maximise Z = 17.5x + 7y subject to the constraints 1x + 3y ≤ 12, 3x+ 1y ≤ 12, x, y > 0

Maximum value of Z is 73.50 at B(3, 3)
Hence the manufacturer must produce 3 packages of nuts and 3 packages of bolts to maximise his profit. The maximum profit is ₹ 73.50.

Maths Chapter 12 Exercise 12.2 Question 5.
A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of ₹ 7 and screws B at a profit of ₹ 10. Assuming that he can sell all the screws he manufactures, how many pack¬ages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
Solution:
Let x be the number of packages of screw A and y be that of screw B.
We have the data as

Automatic machine constraint: 4x + 6y ≤ 240
Hand operated machine constraint: 6x + 3y ≤ 240
Profit function : Z = 7x + 10y
The L.P.P is maximise Z = 1x + 10y subject to the constraints 4x + 6y ≤ 240, 6x + 3y ≤ 240, x, y ≥ 0.

Maximum value of Z is 410 at B(30, 20).
Hence the factory should manufacture 30 packages of screw A and 20 packages of screw B to maximise the profit. The maxi-mum profit is ₹ 410.

Class 12 Chapter 12 Maths Ncert Solutions Question 6.
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting ma-chine for at the most 12 hours. The profit from the sale of a lamp is ₹ 5 and that from a shade is ₹ 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
Solution:
Let the manufacturer produce x lamps and v wooden shades.

Grinding/cutting constraint: 2x + y ≤ 12
Sprayer constraint: 3x + 2y ≤ 20
Profit function : Z = 5x + 3y
∴ The L.P.P is maximise Z = 5x + 3y subject to the constraints 2x + y ≤ 12, 3x + 2y ≤ 20, x, y ≥ 0.

The maximum value of Z is 32 at B(4, 4). Hence to get maximum profit 4 pedestal lamps and 4 wooden shades are to be manufactured. The maximum profit obtained is ₹ 32.

Exercise 12.2 NCERT Solutions Question 7.
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should be company manufacture in order to maximise the profit?
Solution:
Let x be the number of souvenirs of type A and y be the number of souvenirs of type B to be manufactured.
We have the data as

Cutting constraint: 5x+ 8y ≤ 200
Assembling constraint: 10x + 8y ≤ 240
Profit function : Z = 5x + 6y
Then the L.P.P is maximise Z = 5x + 6y subject to the constraints 5x + 8y ≤ 200, 10x + 8y ≤ 240, x, y ≥ 0.

The maximum value of Z is 160 at B(8,20).
Hence for maximum profit 8 souvenirs of type A and 20 souvenirs of type B should be manufactured.
The maximum profit is ₹ 160.

Exercise 12.2 Class 12 NCERT Solutions Question 8.
A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost ₹ 25000 and ₹ 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ₹ 70 lakhs and if his profit on the desktop model is ₹ 4500 and on portable model is ₹ 5000.
Solution:
Let the merchant plan to stock x desktop computers and y portable models.
We have the data as

Demand constraint: x + y ≤ 250
Cost constraint:
25000 x + 40000 y ≤ 7000000
i.e., 5x + 8 ≤ 1400
Profit function : Z = 4500x + 5000y
The L.P.P is maximise Z = 4500x + 5000y
subject to the constraints x + y ≤ 250, 5x + 8y ≤ 1400, x, y ≥ 0.

Maximum value of Z is 1150000 at B(200,50) Hence 200 desktop models and 50 portable models are to be stored to get maximum profit.
The maximum profit is ₹ 1150000.

12.2 Class 12 NCERT Solutions Question 9.
A diet is to contain atlest 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs ₹ 4 per unit food and F₂ costs ₹ 6 per unit. One unit of food F₁ contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Solution:
Let there be x units of food F₁ and y units of food F₂.

Vitamin A constraint: 3x + 6v ≥ 80
Minarals constraint: 4x + 3y ≥ 100
Cost function is : Z = 4x + 6y
The L.P.P is minimise Z = 4x + 6y subject to the constraints 3x + 6v ≥ 80,
4x + 3y ≥ 100, x,y ≥0.

From the table the minimum value of Z is 104. Since the region is unbounded, 104 may or may not be the minimum value of Z.
Consider the inequality 4x + 6y ≤ 104. This half plane has no point common with the feasible region. Hence the minimum value of Z is 104 at B (24, \(\frac { 4 }{ 3 }\))
∴ Minimum cost for diet is ₹ 104

Question 10.
There are two types of fertilisers F₁ and F₂F₁ consists of 10% nitrogen and 6% phosphoric acid and F₂ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that he needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁, costs ₹ 6/kg and F, costs ₹5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Solution:
Let x kg of fertilizer F₁ and y kg of fertilizer
F₂ be used.
We have the data as

Nitrogen constraint: 0.1 x + 0.05 y ≥ 14
i.e., 2x + y ≥ 280
Phosphoric acid constraint: 0.06x + 0.1 y ≥ 14
3x + 5y ≥ 700
Cost function : Z = 6x +5y
The L.P.P is minimise Z = 6x + 5y subject to the constraints 2x + y ≥ 280 and 3x + 5y ≥ 700, x ≥ 0, y ≥ 0

From the table the minimum value of Z is 1000. Since the region is unbounded 1000 may or may not be the minimum value of Z.
Consider the inequality 6x + 5y ≤ 1000
This half plane has no point common with the feasible region
∴ Minimum value of Z is 1000 at B( 100,80).
Hence 100 kg of fertilizer F₁ and 80 kg of fertilizer F₂ is used. The minimum cost of these fertilizers is ₹ 1000.

Question 11.
The corner points of the feasible region de-termined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where р, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0,5) is
a. p = q
b. p = 2q
с. p = 3q
d. q – 3p
Solution:
d. q – 3p
Since maximum Z occurs at (3, 4) and (0, 5), we get Z = 3p + 4q and Z = 0 + 5q
Since both values are same.
∴ 3p + 4q = 5q Hence 3p = q

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise

Question 1.
For each of the differential equations given in Exercises 1 to 12, find the general solution.
i. \(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x\)
ii. \(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x\)
iii. \(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0\)
Solution:
i. \(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x\)
∴ Order 2 and degree = 1

ii. \(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x\)
∴ Order 1 and degree = 3

iii. \(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0\)
∴ Order 4 and degree not defined.

Question 2.
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
i. y = a e^x + be^-x + x²:
\(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}\) – xy + x² – 2 = 0

ii. y = e^x(a cos x + b sin x) :
\(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}\) + 2y = 0

iii. y = x sin 3x :
\(\frac{d^{2} y}{d x^{2}}\) + 9y – 6 cos 3x = 0

iv. x² = 2y² log y :
(x² + y²) \(\frac { dy }{ dx }\) – xy = 0
Solution:
i. y = a e^x + be^-x + x²:
Differentiating w.r.t. x, we get

ii. y = e^x(a cos x + b sin x) :
Differentiating w.r.t. x, we get

iii. y = x sin 3x :
Differentiating w.r.t. x, we get
\(\frac { dy }{ dx }\) = 3x cos3x + sin 3x
\(\frac{d^{2} y}{d x^{2}}\) = 3x(- 3sin 3x) + 3cos 3x + 3cos 3x
= – 9x sin 3x + 6cos 3x
\(\frac{d^{2} y}{d x^{2}}\) + 9y – 6cos 3x
= – 9x sin 3x + 6cos 3x + 9x sin 3x – 6cos 3x = 0
∴ y = xsin 3x is the solution of the differential equation \(\frac{d^{2} y}{d x^{2}}\) + 9y – 6cos 3x = 0

iv. x² = 2y² log y :
Differentiating w.r.t. x, we get

Question 3.
Form the differential equation representing the family of curves given by (x – a)² + 2y² = a², where a is an arbitrary constant.
Solution:
(x – a)² + 2y² = a² … (1)
Differentiating w.r.t. x, we get
2(x – a) + 4yy’ = 0
x – a + 2yy’ = 0
∴ a = x + 2yy’
(1) → (- 2yy’)² + 2y² =[x + 2yy’]²
4(yy’)² + 2y² = x² + 4 xyy’ + 4 (yy’)²
2y² = x² + 4xyy’
∴ y’ = \(\frac{2 y^{2}-x^{2}}{4 x y}\) is the required differential equation.

Question 4.
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:

x² – y² = C(x² + y²)² where C = C²₁ is the general solution.

Question 5.
Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
Solution:

The equation of the required circles is
(x – r)² + (y – r)² = r² … (1)
x² + y² – 2rx – 2ry + r² = 0
Differentiating (1) w.r.t. x, we get
2x + 2yy₁ – 2r – 2ry₁ = 0
r = \(\frac{x+y y_{1}}{1+y_{1}}\)
Substituting r in (1), we get

is the required differential equation.

Question 6.
Find the general solution of the differential equation \(\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0\).
Solution:

∴ sin^-1y + sin^-1x = C, is the general solution.

Question 7.
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 the parameter.
Solution:

Question 8.
Find the equation of the curve passing through the point (0, \(\frac { π }{ 4 }\)) whose differential equation is sin x cos y dx + cos x siny dy = 0.
Solution:
sin x cos y dx + cos x sin y dy = 0.
Divide by cos x cos y, we get
\(\frac{\sin x}{\cos x} d x+\frac{\sin y}{\cos y} d y=0\)
tan x dx + tan y dy = 0
Integrating both sides, we get
∫tan x dx + ∫tan y dy = ∫o dx
log|sec x| + log|sec y| = log|C|
log|sec x secy| = log|C|
sec x sec y = C … (1)
(1) Passes through the point
we get C = \(\sqrt{2}\)
∴ (1) → sec x secy = \(\sqrt{2}\)
⇒ cos y = \(\frac{\sec x}{\sqrt{2}}\), is the equation of the curve.

Question 9.
Find the particular solution of the differential equation
(1 + e^2x) dy + (1 + y²)e^x dx = 0. Given that y = 1 when x = 0.
Solution:
(1 + e^2x) dy + (1 + y²)e^x dx = 0
Divide by (1 + e^2x)(1 + y²), we get

Question 10.
Solve the differential equation
\(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y\) (y ≠ 0).
Solution:

i.e., \(e^{\frac{x}{y}}\) = y + C is the general solution.

Question 11.
Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy. Given that y = – 1, when x = 0.
Solution:
(x – y)(dx + dy) = dx – dy
xdx – ydx + xdy – ydy = dx – dy
(x – y – 1)dx = (y – x – 1)dy

t + log|t| = 2x + C
x – y + log|x – y|= 2x + C … (2)
When x = 0, y = – 1
∴ (2) → 1 + log 1 = 0 + C
C = 1
∴ (2) → x – y + log |x – y| = 2x + 1
Hence x + y + 1 = log|x – y| is the particular solution

Question 12.
Solve the differential equation \(\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1(x \neq 0)\).
Solution:

i.e., \(y e^{2 \sqrt{x}}=2 \sqrt{x}+C\) is the general solution.

Question 13.
Find a particular solution of the differential equation \(\frac { dy }{ dx }\) + y cot x = 4x cosec x (x ≠ 0).
Given that y = 0 when x = \(\frac { π }{ 2 }\)
Solution:
\(\frac { dy }{ dx }\) + y cot x = 4x cosec x is a linear differential equation.
∴ P = cot x and Q = 4x cosec x
∫Pdx = ∫cot x dx = log sinx
∴ I.F. = e^{∫p dx} = e^{log sinx} = sin x
The solution is y(I.F) = ∫Q(I.F)dx + C
y sinx = ∫4x cosec x sin xdx + C
= ∫4x dx + C
y sin x = 2x + C … (1)
When x = \(\frac { π }{ 2 }\), y = 0
(1) → o = \(\frac { π² }{ 2 }\) + C
∴ C = \(\frac { π² }{ 2 }\)
Hence the particular solution is
ysin x = 2x² – \(\frac { π² }{ 2 }\)

Question 14.
Find a particular solution of the differential equation (x + 1) \(\frac { dy }{ dx }\) = 2e^-y – 1, given that y = 0 when x = 0.
Solution:

Hence y = log\(\left|\frac{2 x+1}{x+1}\right|\), is the particular solution.

Question 15.
The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?
Solution:
Let P be the population at time t
Given \(\frac { dP }{ dt }\) ∝ P
i.e, \(\frac { dP }{ dt }\) = kP, k is a constant
i.e., \(\frac { dP }{ dt }\) = kdt
Integrating both sides, we get
∫\(\frac { dP }{ dt }\) = k ∫dt
log|P| = kt + C … (1)
The initial population in 1999 is 20,000
i.e., when t = 0, P = 20,000
Hence (1) → log 20,000 = k(0) + C
∴ C = log 20,000
After 5 years, i.e., in 2004, the population is 25,000
i.e., when t = 5, P = 25000
Hence (1) → log 25000 = 5k + log 20000
log 25000 – log 20000 = 5k

Question 16.
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:
c. y = Cx
\(\frac{y d x-x d y}{y}\) = 0
∴ ydx – xdy = 0
ydx = xdy
\(\frac { dy }{ y }\) = \(\frac { dx }{ x }\)
Integrating both sides, we get
∫\(\frac { dy }{ y }\) = ∫\(\frac { dx }{ x }\)
log|y| = log|x| + log|C|
log|y| = log|Cx|
y = Cx

Question 17.
The general solution of a differential equation of the type \(\frac { dx }{ dy }\) + P₁, x = Q₁ is

Solution:
\(e^{\int {P}_{1} d y}\) is the integrating factor
∴ The general solution is
\(x \cdot e^{\int {P}_{1} d y}=\int\left({Q}_{1} e^{\int {P}_{1} d y}\right) d y+ {C}\)

Question 18.
The general solution of the differential equation e^xdy + (ye^x + 2x) dx = 0 is
a. x e^y + x² = C
b. x e^y + y² =C
c. y e^x + x² = C
d. y e^y + x² = C
Solution:

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3

Question 1.
\(\frac{x}{a}+\frac{y}{b}\) = 1
Solution:
\(\frac{x}{a}+\frac{y}{b}\) = 1
i.e., bx + ay = ab … (1)
Differentiating (1) w.r.t. x, we get
b + ay’ = 0
i.e., ay’ = – b
y = \(\frac { – b }{ a }\)
Differentiating again w.r.t. x, we get y’ = 0
which is the required differential equation.

Question 2.
y² = a(b² – x²)
Solution:
y² = a(b² – x²)
i.e., y² = ab² – x²a
Differentiating w.r.t. x, we get 2yy’ = – 2ax
i.e., yy’ = – ax
\(\frac { yy’ }{ x }\) = – a
Differentiating again w.r.t. x, we get
\(\frac{x\left[y y^{\prime \prime}+\left(y^{\prime}\right)^{2}\right]-y y^{\prime}}{x^{2}}\) = 0
i.e., x yy” + x(y’)² – yy’ = 0
which is the required differential equation.

Question 3.
y = ae^3x + be^-2x
Solution:
Given that
y = ae^3x + be^-2x … (i)
Differentiating w.r.t. x two times, we get
y’ = 3ae^3x – 2be^-2x
y” = 9ae^3x + 4be^-2x
i.e., y” = 6ae^3x + 3ae^3x + 6be^-2x – 2be^-2x
= 3ae^3x – 2be^-2x + 6(ae^3x + be^-2x)
= y’ + 6y
i.e., y” – y’ – 6y = 0 is the required differential equation.

Question 4.
y = e^2x (a + bx)
Solution:
y = e^2x (a + bx)
y’ = be^2x + (a + bx)e^2x x 2
y’ = be^2x + 2y
Differentiating again w.r.t. x, we get
y” = 2be^2x + 2y’
i.e., y” = 2(y’ – 2y) + 2y’ [∵ y’ – 2y = be^2x]
y” = 4y’ – 4y
y” – 4y’ + 4y = 0 is the required differential equation.

Question 5.
y = e^x(a cosx + b sinx)
Solution:
The curve y = e^x(a cosx+b sinx) …(i)
Differentiating (1) w.r.t. x, we get
y’ = e^x(- a sinx + b cosx) + e^x(a cosx + b sinx)
y’ = e^x(b cos x – a sin y) + y
i.e., y’ – y = e^x(b cosx – a sinx) … (2)
Differentiating (2) w.r.t. x, we get
y”- y’ = – e^x(b sinx + a cosx) + e^x(b cosx – a sinx)
y” – y’ = – y + y’- y [from (1) and (2)]
y” – 2y’ + 2y = 0 is the required differential equation.

Question 6.
Form the differential equation of the family of circles touching the y axis at origin
Solution:

Circles touching y axis at the orgin will have its centre on x-axis and will pass through the origin.
So the centre of circle will be (a, 0) and radius a.
∴ Equation of the circle is (x – a)² + (y – 0)² = a²
i.e., x² + y² – 2ax = 0
i.e., \(\frac{x^{2}+y^{2}}{x}\) = 2a
\(\frac{x\left(2 x+2 y y_{1}\right)-\left(x^{2}+y^{2}\right)}{x^{2}}\) = 0
Differentiating w.r.t. x, we get
2x² + 2xyy₁ – x² – y² = 0
x² – y² + 2xyy₁ = 0 or y² – x² – 2xyy₁ = 0

Question 7.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solution:

Consider the family of parabolas having focus (0, a) at the positive y-axis, where a is an arbitrary constant.
∴ The equation of family of parabolas is x² = 4ay … (1)
Differentiating both sides w.r.t. x, we get
2x = 4 ay’
i.e., 4a = \(\frac { 2x }{ y’ }\) … (2)
Substituting (2) in (1) we get x² = (\(\frac { 2x }{ y’ }\))y
x²y’ – 2xy = 0
i.e., xy’ – 2y = 0 is the required differential equation.

Question 8.
Form the differential equation of family of ellipses having foci on y-axis and centre at origin.
Solution:
The equation of family ellipses having foci at y- axis is

Question 9.
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Solution:
The equation of family of hyperbolas having foci on x axis is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, a and b are the parameters.

i.e., xyy” + x(y’)² – yy’ = 0 is the required differential equation.

Question 10.
Form the differential equation of the family of circles having centre on y-axis and radius 3 units
Solution:
Consider a circle of radius 3 unit and centre on (0, a). The equation of the circle is
(x – 0)² + (y – a)² = 3²
x² + (y – a)² = 9 … (1)
Differentiating both sides w.r.t. x, we get
2x + 2(y – a)y’ = 0
x + yy’ – ay’ = 0
ay’ = x + yy’

⇒ (x² – 9)(y’)² + x² = 0 is the required differential equation.

Question 11.
Which of the following differential equation has y = \({ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }\) as the general solution?
(a) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)
(c) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +1=0\)
(d) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -1=0\)
Solution:
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)

Question 12.
Which of the following differential equations has y = x as one of its particular solution ?
(a) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=x\)
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=x\)
(c) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=0\)
(d) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=0\)
Solution:
(c) y = x

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-3/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3

Class 12 Maths Ncert Solutions Chapter 10.3 Question 1.
Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes \(\sqrt{3}\) and 2 respectively, and such that \(\vec{a}\).\(\vec{b}\) = \(\sqrt{6}\)
Solution:
|\(\vec{a}\)|= \(\sqrt{3}\)
and \(\vec{b}\) = 2, \(\vec{a}\).\(\vec{b}\) = \(\sqrt{6}\)
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{\sqrt{6}}{(\sqrt{3})(2)}=\frac{1}{\sqrt{2}}\)
∴ θ = \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)

Question 2.
Find the angle between the vectors \(\hat{i}-2 \hat{j}+3 \hat{k} \text { and } 3 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:
Let \(\vec{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}\) = \(\hat{3i}-2 \hat{j}+ \hat{k}\)
\(\vec{a}\).\(\vec{b}\) = \((\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+\hat{k})\)
= 1(3) + (- 2)(- 2) + 3(1) = 3 + 4 + 3 = 10

Question 3.
Find the projection of the vector \(\overrightarrow { i } -\overrightarrow { j }\), on the line represented by the vector \(\overrightarrow { i } +\overrightarrow { j }\).
Solution:

Question 4.
Find the projection of the vector \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(7 \hat{i}-\hat{j}+8 \hat{k}\)
Solution:

Question 5.
Show that each of the given three vectors is a unit vector \(\frac { 1 }{ 7 } \left( 2\hat { i } +3\hat { j } +6\hat { k } \right) ,\frac { 1 }{ 7 } \left( 3\hat { i } -6\hat { j } +2\hat { k } \right) ,\frac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } -3\hat { k } \right)\)
Also show that they are mutually perpendicular to each other.
Solution:
Let \(\vec{a}\) = \(\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})\), \(\vec{b}\) = \(\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})\) and \(\vec{c}\) = \(\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\)

Here \(\vec{a}\).\(\vec{b}\) = 0, \(\vec{b}\).\(\vec{c}\) = 0 and \(\vec{a}\).\(\vec{c}\) = 0
∴ The vectors \(\vec{a}\).\(\vec{b}\) and \(\vec{c}\) are mutually per-pendicular vectors.

Question 6.
\(Find\left| \overrightarrow { a } \right| and\left| \overrightarrow { b } \right| if\left( \overrightarrow { a } +\overrightarrow { b } \right) \cdot \left( \overrightarrow { a } -\overrightarrow { b } \right) =8\quad and\left| \overrightarrow { a } \right| =8\left| \overrightarrow { b } \right| \)
Solution:

Question 7.
Evaluate the product :
\((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\)
Solution:
\(\left( 3\overrightarrow { a } -5\overrightarrow { b } \right) \cdot \left( 2\overrightarrow { a } +7\overrightarrow { b } \right) \)
\(=6\overrightarrow { a } .\overrightarrow { a } -10\overrightarrow { b } \overrightarrow { a } +21\overrightarrow { a } .\overrightarrow { b } -35\overrightarrow { b } .\overrightarrow { b } \)
\(=6{ \left| \overrightarrow { a } \right| }^{ 2 }-11\overrightarrow { a } \overrightarrow { b } -35{ \left| \overrightarrow { b } \right| }^{ 2 }\)

Question 8.
Find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\) having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac { 1 }{ 2 }\)
Solution:

Question 9.
Find |\(\vec{a}\)| , if for a unit vector \(\vec{a}\), \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 12
Solution:

Question 10.
If \(\overrightarrow { a } =2\hat { i } +2\hat { j } +3\hat { k } ,\overrightarrow { b } =-\hat { i } +2\hat { j } +\hat { k } and \overrightarrow { c } =3\hat { i } +\hat { j } \) such that \(\overrightarrow { a } +\lambda \overrightarrow { b } \bot \overrightarrow { c } \) , then find the value of λ.
Solution:

Question 11.
Show that \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a} \), is per-pendicular to \(|\vec{a}| \vec{b}-|\vec{b}| \vec{a} \) for any two non-zero vectors \(\vec{a} \text { and } \vec{b}\)
Solution:

Question 12.
If \(\overrightarrow { a } \cdot \overrightarrow { a } =0\quad and\quad \overrightarrow { a } \cdot \overrightarrow { b } =0\), then what can be concluded about the vector \(\overrightarrow { b } \) ?
Solution:
\(\vec{a}\).\(\vec{a}\) = 0 ⇒ \(\vec{a}\) is a zero vector.
since \(\vec{a}\) = \(\vec{0}\), \(\vec{a}\).\(\vec{b}\) = 0 for any vector \(\vec{b}\)
Vector \(\vec{b}\) be any vector.

Question 13.
If \(\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } \) are the unit vector such that \(\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =0\) , then find the value of \(\overrightarrow { a } .\overrightarrow { b } +\overrightarrow { b } .\overrightarrow { c } +\overrightarrow { c } .\overrightarrow { a } \)
Solution:

Question 14.
If either vector \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}\), then \(\vec{a} \cdot \vec{b}\). But the converse need not be true. Justify your answer with an example.
Solution:

Thus two non-zero vectors \(\vec{a}\) and \(\vec{b}\) may have \(\vec{a}\).\(\vec{a}\) zero.

Question 15.
If the vertices A,B,C of a triangle ABC are (1, 2, 3) (-1, 0, 0), (0, 1, 2) respectively, then find ∠ABC.
Solution:

Question 16.
Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are collinear.
Solution:
A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are the points.

∴ \(\overrightarrow{AB}\)\(\overrightarrow{AC}\) are parallel and A is a common point. Therefore A, B, C are collinear.

Question 17.
Show that the vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k} \text { and } \quad 3 \hat{i}-4 \hat{j}-4 \hat{k}\) from the vertices of a right angled triangle.
Solution:
Let A, B and C be the vertices of the triangle with the vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k} \text { and } \quad 3 \hat{i}-4 \hat{j}-4 \hat{k}\)
∴ \(\overrightarrow{AB}\) = p.v. of B – p.v. of A

∴ A, B, C are the vertices of ∆ ABC
\(\overrightarrow{BC}\).\(\overrightarrow{CA}\) = (2)(-1) + (-1)(3) + (1)(5)
= – 2 – 3 + 5 = 0
∴ \(\overrightarrow{BC}\)⊥\(\overrightarrow{CA}\)
Hence triangle ABC is a right triangle

Question 18.
If \(\overrightarrow { a } \) is a non-zero vector of magnitude ‘a’ and λ is a non- zero scalar, then λ \(\overrightarrow { a } \) is unit vector if
(a) λ = 1
(b) λ = – 1
(c) a = |λ|
(d) a = \(\frac { 1 }{ \left| \lambda \right| } \)
Solution:
\(\left| \overrightarrow { a } \right| =a\)
Given : \(\lambda \overrightarrow { a } \) is a unit vectors.
\(|\lambda \vec{a}|=1 \Rightarrow|\lambda||\vec{a}|=1 \Rightarrow|\lambda| a=1 \Rightarrow a=\frac{1}{|\lambda|}\)

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-2/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.2

Ex 9.2 Class 12 NCERT Solutions Question 1.
y = e^x + 1 : y”- y’ = 0
Solution:
y = e^x + 1
Differentiating w.r.t. x, we get
y’ = e^x
i.e., y” = y’
⇒ y” – y’ = 0
Hence y = e^x is a solution of y” – y’ = 0.

Ex 9.2 Class 12 Maths Ncert Solutions Question 2.
y = x² + 2x + C : y’ – 2x – 2 = 0
Solution:
y = x² + 2x + C
Differentiating w.r.t. x, we get y’ = 2x + 2
i.e., y’ – 2x – 2 = 0
Hence y = x² + 2x + C is a solution of y’ – 2x – 2 = 0

Question 3.
y = cos x + C : y’ + sin x = 0
Solution:
y = cos x + C
Differentiating w.r.t., x, we get y’ = – sin x
i.e., y’ + sin x = 0
Hence y = cos x + C is a solution of y’ + sin x = 0

Question 4.
y = \(\sqrt{1+x^{2}} \quad: \quad y^{\prime}=\frac{x y}{1+x^{2}}\)
Solution:
\(\sqrt{1+x^{2}}\)
Differentiating w.r.t. x, we get

Question 5.
y – Ax : xy’ = y (x ≠ 0)
Solution:
y = Ax … (1)
Differentiating w.r.t., x, we get y’ = A
i.e., y’ = \(\frac { y }{ x }\) [From(1), A = \(\frac { y }{ x }\)]
i.e., xy’ = y (x ≠ 0)
∴ y = Ax is a solution of xy’ = y.

Question 6.
y = x sin x : xy’= y + x \(\sqrt{x^{2}-y^{2}}\)
(x ≠ 0 and x > y or x < – y)
Solution:
y = x sinx … (1)
Differentiating (1) w.r.t. x, we get
y’ = x cos x + sin x … (2)
From (1) sin x = \(\frac { y }{ x }\)

∴ y = x sinx is a solution of xy’ = x\(\sqrt{x^{2}-y^{2}}\) + y

Question 7.
Ay = logy + C : y’ = \(\frac{y^{2}}{1-x y}(x y \neq 1)\)
Solution:
xy = logy + C
Differentiating w.r.t. x, we get
xy’ + y = \(\frac { 1 }{ y }\) y’
i.e., xyy’ + y² = y’
y’ – xyy = y²
y'(1 – xy) = y²
i.e., y’ = \(\frac{y^{2}}{1-x y}\)
Hence xy = logy + C is a solution of y’ = \(\frac{y^{2}}{1-x y}\)

Question 8.
y – cos y = x : (ysiny + cosy + x)y’ = y.
Solution:
y – cosy = x … (1)
Differentiating (1) w.r.t. x, we get
y’ + sin y . y’ = 1
i. e., yy’ + yy’sin y = y
(x + cos y)y’ + yy’sin y = y
[since from (1) y = x + cosy]
xy’ + y’cosy+ yy’siny = y
(y sin y + cos y+ x)y’ = y
Hence y – cosy = x is a solution of
(y sin y + cos y + x)y’ = y

Question 9.
x + y = tan^-1 y : y²y’ + y² + 1 = 0
Solution:
x + y = tan^-1
Differentiating w.r.t. x, we get,
1 + y’ = \(\frac{1}{1+y^{2}}\)
i.e., (1 + y²)(1 + y’) = y ‘
1 + y’ + y² + y²y’ = y’
i.e., y²y’ + y² + 1 = 0
Hence x + y = tan^-1 is a solution of y²y’ + y² + 1 = 0

Question 10.
\(y=\sqrt { { a }^{ 2 }-{ x }^{ 2 } } x\in (-a,a);x+y\frac { dy }{ dx } =0,(y\neq 0)\)
Solution:
y = \(\sqrt{a^{2}-x^{2}}\) , x ∈ (- a, a)
Differentiating w.r.t. x, we get

Hence y = \(\sqrt{a^{2}-x^{2}}\) is a solution of x + y\(\frac { dy }{ dx }\) = 0

Question 11.
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:
d. 4
The general solution of a differential equation contains as many arbitrary constants as the order of the differential equation. Since the differential equation is of order 4, its solution contains 4 arbitrary constants.

Question 12.
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:
d. 0
The particular solution is free from the arbitrary constants.