Determine order and degree (if defined) of differential equations given in Exercises 1 to 10 .
1.
Solution: We have,
The highest order derivative present in the differential equation is , so its order is four.
The differential equation is not a polynomial equation. Therefore, its degree is not defined .
2.
Solution: We have, y'+5y=0
The highest order derivative present in the differential equation is y’ , so its order is one.
The differential equation is a polynomial equation. Therefore, its degree is one .
3.
Solution: We have,
The highest order derivative present in the differential equation is , so its order is two.
The degree of the differential equation is 1 .
4.
Solution: We have,
The highest order derivative present in the differential equation is , so its order is two .
The differential equation is not a polynomial equation. Therefore, its degree is not defined .
5.
Solution: We have,
The highest order derivative present in the differential equation is , so its order is two .
The differential equation is a polynomial equation. Therefore, its degree is one .
6.
Solution: We have,
The highest order derivative present in the differential equation is , so its order is two .
The differential equation is a polynomial equation. Therefore, its degree is two .
7.
Solution: We have,
The highest order derivative present in the differential equation is ' , so its order is three .
The differential equation is a polynomial equation in . Therefore, its degree is one .
8.
Solution: We have,
The highest order derivative present in the differential equation is , so its order is two .
The differential equation is a polynomial equation in (power is 1). Therefore, its degree is one .
9.
Solution: We have,
The highest order derivative present in the differential equation is , so its order is two .
The differential equation is a polynomial equation in (power is 1). Therefore, its degree is one .
10.
Solution: We have,
The highest order derivative present in the differential equation is , so its order is two .
The differential equation is a polynomial equation in (power is 1). Therefore, its degree is one .
11. The degree of the differential equation is :
(A) 3 (B) (C) 1 (D) not defined
Solution: We have,
The differential equation is not a polynomial equation . Therefore, its degree is not defined .
The correct answer (D) not defined .
12. The order of the differential equation is
(A) 2 (B) (C) 0 (D) not defined
Solution: We have,
The highest order derivative present in the differential equation is , so its order is two .
The correct answer (A) 2.
In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equations :
1.
Solution: We have,
Differentiating with respect to , we get
Again,
.
Therefore, the given functions is a solution of the corresponding differential equations .
2.
Solution: We have,
Differentiating with respect to , we get
Therefore, the given functions is a solution of the corresponding differential equations .
3.
Solution: We have,
Differentiating with respect to x , we get
Therefore, the given functions is a solution of the corresponding differential equations .
4.
Solution: We have,
Differentiating with respect to x , we get
Therefore, the given functions is a solution of the corresponding differential equations .
5.
Solution: We have, ……..(i)
Differentiating with respect to x , we get
Therefore, the given functions is a solution of the corresponding differential equations .
6.
Solution: We have,
And
Differentiating with respect to x , we get
Therefore, the given functions is a solution of the corresponding differential equations .
7.
Solution: We have,
Differentiating with respect to , we get
Therefore, the given functions is a solution of the corresponding differential equations .
8.
Solution: We have,
Differentiating with respect to , we get
LH.S. :
R.H.S.
Therefore, the given functions is a solution of the corresponding differential equations .
9.
Solution: We have,
Differentiating with respect to x , we get
Therefore, the given functions is a solution of the corresponding differential equations .
10.
Solution: We have,
Differentiating with respect to x , we get
Therefore, the given functions is a solution of the corresponding differential equations .
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: The correct Answer (D) 4 .
The number of arbitrary constants in the general solution of a differential equation of fourth order depends on the characteristics of the equation itself. So, for a fourth-order differential equation, the number of arbitrary constants in the general solution is 4.
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: The correct answer (D) 0 .
The number of arbitrary constants in the particular solution of a differential equation of any order is always 0.
For each of the differential equations in Exercise 1 to 10 , find the general solution :
1.
Solution: We have,
This is the required general solutions .
2.
Solution: We have,
This is the required general solutions .
3.
3.
Solution : We have,
[Where ]
This is the required general solutions .
4.
Solution: We have,
This is the required general solutions .
5.
Solution: We have,
This is the required general solutions .
6.
Solution: We have,
This is the required general solutions .
7.
Solution: We have,
Let
[ Where is the arbitrary constant ]
This is the required general solutions .
8.
Solution: We have,
This is the required general solutions .
9.
Solution : We have,
This is the required general solutions .
10.
Solution: We have,
This is the required general solutions .
For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:
11. ; when
Solution: We have,
Let
Putting , then
Putting x=0 , then
Putting , then
If and , then
Therefore,
12. ; when
Solution: We have,
Let ,
Putting , then
Putting , then
Putting , then
If and , then
Therefore,
13. ; when
Solution: We have,
If and , then
Therefore,
14. ; when
Solution: We have,
If and , then
15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is .
Solution: We have,
Let,
Now ,
If and then
16. For the differential equation , find the solution curve passing through the point (1, –1).
Solution: We have,
If and , then
Therefore,
17. 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: The slope of the tangent is .
The product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
A/Q ,
If and , then
Therefore, the equation of a curve is .
18. At any point () 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: We know that, the slope of the tangent
Therefore, the slope of the tangent
A/Q,
If and , then
Therefore,
So, the equation of the curve is .
19. 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: We know that , the volume of the spherical balloon
Let be the radius of the spherical balloon .
A/Q, [Where M is constant .]
If and , then
If and , then
Therefore, the radius of the balloon after t seconds is unit .
20. In a bank, principal increases continuously at the rate of per year. Find the value of if Rs 100 double itself in 10 years ( ).
Solution: Let P be the principal ant time be t .
A/Q,
If and , then
If and , then
Therefore, the value of is 6.931 % .
21. 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 ( 1.648).
Solution: Let P be the principal ant time be t .
A/Q,
If and , then
If , then
Therefore, the amount will worth Rs 1648 after 10 years .
22. 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: Let be the number of bacteria count at the end of t hours , then
If and , then
If , , then
Putting in (i), then
Required time is
23. The general solution of the differential equation is
(A) (B)
(C) (D)
Solution: We have,
The correct Answer (A)
In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:
11. when x .
12. when.
13. ; when
14. , when
15. ; when
16. A homogeneous differential equation of the from can be solved by making the substitution.
(A) (B) (C) (D)
17. Which of the following is a homogeneous differential equation?
(A)
(B)
(C)
(D)
For each of the differential equations given in Exercises 1 to 12, find the general solution:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition:
13. when
14. when x .
15. when .
16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point () is equal to the sum of the coordinates of the point.
17. 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.
18. The Integrating Factor of the differential equation is :
(A) (B) (C) (D)
19. The Integrating Factor of the differential equation is
(A) (B) (C) (D)
1. For each of the differential equations given below, indicate its order and degree (if defined).
(i) (ii) (iii)
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)
(ii)
(iii)
(iv)
4. Prove that is the general solution of differential equation ,where is a parameter.
5. Find the general solution of the differential equation .
6. Show that the general solution of the differential equation is given by , where A is parameter.
7. Find the equation of the curve passing through the point whose differential equation is .
8. Find the particular solution of the differential equation , given that when x .
9. Solve the differential equation .
10. Find a particular solution of the differential equation , given that , when . (Hint: put )
11. Solve the differential equation .
12. Find a particular solution of the differential equation ,given that when x .
13. Find a particular solution of the differential equation , given that when .
14. The general solution of the differential equation is
(A) (B) (C) (D)
15. The general solution of a differential equation of the type is :
(A)
(B)
(C)
(D)
16. The general solution of the differential equation is
(A) (B) (C) (D)