Question 1.
In each of the following examples verify that the given expression is a solution of the corresponding differential equation.
Solution:
xy = log y + c
Differentiating w.r.t. x, we get
![Chapter 6 Differential Equations Ex 6.3 1 word image 21266 2](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-2.jpeg)
Hence, xy = log y + c is a solution of the D.E.
![Chapter 6 Differential Equations Ex 6.3 2 word image 21266 3](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-3.png)
Differentiating w.r.t. x, we get
Differentiating again w.r.t. x, we get
![Chapter 6 Differential Equations Ex 6.3 6 word image 21266 7](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-7.png)
![Chapter 6 Differential Equations Ex 6.3 7 word image 21266 8](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-8.jpeg)
![Chapter 6 Differential Equations Ex 6.3 8 word image 21266 9](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-9.png)
![Chapter 6 Differential Equations Ex 6.3 9 word image 21266 10](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-10.jpeg)
![Chapter 6 Differential Equations Ex 6.3 10 word image 21266 11](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-11.png)
Differentiating w.r.t. x, we get
Differentiating again w.r.t. x, we get
![Chapter 6 Differential Equations Ex 6.3 15 word image 21266 16](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-16.png)
![Chapter 6 Differential Equations Ex 6.3 16 word image 21266 17](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-17.jpeg)
Solution:
This is the general solution.
![Chapter 6 Differential Equations Ex 6.3 18 word image 21266 21](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-21.png)
![Chapter 6 Differential Equations Ex 6.3 19 word image 21266 22](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-22.png)
![Chapter 6 Differential Equations Ex 6.3 20 word image 21266 23](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-23.png)
Multiplying throughout by 4, this becomes
Solution:
Solution:
Solution:
This is the general solution.
Question 3.
For each of the following differential equations, find the particular solution satisfying the given condition:
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
![Chapter 6 Differential Equations Ex 6.3 33 word image 21266 39](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-39.png)
![Chapter 6 Differential Equations Ex 6.3 34 word image 21266 40](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-40.jpeg)
![Chapter 6 Differential Equations Ex 6.3 35 word image 21266 41](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-41.jpeg)
![Chapter 6 Differential Equations Ex 6.3 36 word image 21266 42](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-42.png)
![Chapter 6 Differential Equations Ex 6.3 37 word image 21266 43](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-43.jpeg)
![Chapter 6 Differential Equations Ex 6.3 38 word image 21266 44](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-44.png)
![Chapter 6 Differential Equations Ex 6.3 39 word image 21266 45](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-45.jpeg)
![Chapter 6 Differential Equations Ex 6.3 40 word image 21266 46](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-46.png)
![Chapter 6 Differential Equations Ex 6.3 41 word image 21266 47](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-21266-47.png)
Question 4.
Reduce each of the following differential equations to the variable separable form and hence solve:
Solution:
Solution:
Solution:
Integrating both sides, we get
Integrating both sides, we get
∫dx = ∫seu du
∴ x = tan u + c
∴ x = tan(x – 2y) + c
This is the general solution.
(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1
Solution:
(2x – 2y + 3) dx – (x – y + 1) dy = 0
∴ (x – y + 1) dy = (2x – 2y + 3) dx
∴ u – log|u + 2| = -x + c
∴ x – y – log|x – y + 2| = -x + c
∴ (2x – y) – log|x – y + 2| = c
This is the general solution.
Now, y = 1, when x = 0.
∴ (0 – 1) – log|0 – 1 + 2| = c
∴ -1 – o = c
∴ c = -1
∴ the particular solution is
(2x – y) – log|x – y + 2| = -1
∴ (2x – y) – log|x – y + 2| + 1 = 0