NCERT Solutions Class 12 Mathematics Chapter 9 Differential Equations

Exercise 9.1

1. Determine order and degree (If defined) of differential equation d⁴y /dx⁴ + sin(y')  = 0 
Solution
d⁴y /dx⁴ + sin(y''')  = 0 
⇒ y''' + sin(y''') = 0 
The highest order derivative present in the differential equation is y''''. Therefore, its order is four. 
The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

2. Determine order and degree(if defined) of differential equation y' + 5y = 0
Solution
The given differential equation is : 
y' + 5y = 0 
The highest order derivative present in the differential equation is y' . Therefore, its order is one. 
It is a polynomial equation in y'. The highest power raised to y' is 1. Hence, its degree is one.

3. Determine order and degree (if defined of differential equation (ds/dt)⁴ + 3s d²s/dt² = 0 
Solution

The highest order derivative present in the given differential equation is d²s/dt². Therefore, its order is two .
It is a polynomial equation in d²s/dt²  and ds/dt. The power raised to d²s/dt² is 1. 
Hence, its degree is one. 

4. Determine order and degree (if defined) of differential equation d²y/(ds²)² + cos (dy/dx) = 0 
Solution

The highest order derivative present in the given differential equation is d² y/dx². Therefore, its order is 2. 
The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined. 

5. Determine order and degree (if defined) of differential equation d² y/dx² = cos 3x + sin 3x 
Solution
d²y/dx² = cos 3x + sin 3x 
⇒ d²y/dx² - cos 3x - sin 3x = 0 
The highest order derivative present in the differential equation is d²y/dx² . Therefore, its order is two. 
It is a polynomial equation in d² y/dx² and the power raised to d²y/dx² is 1. 
Hence, its degree is one. 

6. Determine order and degree (if defined) of differential equation 
(y''') + (y'')³ + (y')⁴ + y⁵ = 0
Solution
(y''') + (y'')³ + (y')⁴ + y⁵ = 0 
The highest order derivative present in the differential equation is y'''. Therefore, its order is three. 
The given differential equation is a polynomial equation in y''' , y'' , and y' . 
The highest power raised to y''' is 2. Hence, its degree is 2. 

7. Determine order and degree(if defined) of differential equation y′′′ + 2y″ + y′ = 0
Solution
y′′′ + 2y″ + y′ = 0
The highest order derivative present in the differential equation is y'''. Therefore, its order is three. 
It is a polynomial equation in y''', y'' and y' . The highest power raised to y''' is 1. Hence, its degree is 1. 

8. Determine order and degree(if defined) of differential equation y′ + y = e^x  .    
Solution
y′ + y = e^x 
⇒ y' + y - e^x  = 0 
The highest order derivative present in the differential equation is y'. Therefore, its order is one. 
The given differential equation is a polynomial equation in y' and the highest power raised to y' is one. Hence, its degree is one. 

9. Determine order and degree (if defined) of differential equation y'' + (y')² + 2y = 0  
Solution
y'' + (y')² + 2y = 0  
The highest order derivative present in the differential equation is y''. Therefore, its order is two. 
The given differential equation is a polynomial equation in y'' and y' and the highest power raised to y'' is one. 
Hence, its degree is one . 

10. Determine order and degree(if defined) of differential equation y″ + 2y′ + sin y = 0
Solution
y'' + 2y' + sin y = 0 
The highest order derivative present in the differential equation is y''. Therefore, its order is two. 
This is a polynomial equation in y'' and y' and the highest power raised to y'' is one. Hence, its degree is one. 

11. The degree of the differential equation 
(d²y/dx² )³ + (dy/dx)² + sin(dy/dx) + 1 = 0 
(A) 3 
(B) 2 
(C) 1 
(D) not defined 
Solution

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined. 
Hence, the correct answer is D.

12. The order of the differential equation  2x²(d²y/dx²) + 3(dy/dx) + y = 0 is 
(A) 2 
(B) 1 
(C) 0 
(D) not defined 
Solution

The highest order derivative present in the given differential equation is d² y/dx² . Therefore, its order is two. 
Hence, the correct answer is A.

Exercise 9.2

1. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation
y = e^x + 1 : y″ – y′ = 0
Solution
y = e^x + 1 
Differentiating both sides of this equation with respect to x, we get : 

⇒ y' = e^x
Substituting the values of y' and y'' in the given differential equation, we get the L.H.S. as : 
y'' - y' = e^x - e^x = 0 R.H.S. 
Thus, the given function is the solution of the corresponding differential equation.

2. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation
y = x² + 2x + C : y′ – 2x – 2 = 0 
Solution
y = x² + 2x + C 
Differentiating both sides of this equation with respect to x, we get : 

⇒ y' = 2x + 2 
Substituting the value of y' in the given differential equation, we get : 
L.H.S = y' - 2x - 2 = 2x + 2 - 2x - 2 = 0 R.H.S. 
Hence, the given function is the solution of the corresponding differential equation.

3. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation
y = cos x + C : y′ + sin x = 0
Solution
y = cos x + C 
Differentiating both sides of this equation with respect to x, we get : 

⇒ y' = - sin x
Substituting the value of y' in the given differential equation, we get : 
L.H.S. = y' + sin x = - sin x + sin x = 0 = R.H.S. 
Hence, the given function is the solution of the corresponding differential equation.

4. Verify that the given functions (explicit of implicit) is a solution of the corresponding differential equation y = √(1 + x²) : y' = xy/(1+ x²)
Solution
y = √(1 + x² )
Differentiating both sides of the equation with respect to x, we get : 

5. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation
y = Ax : xy′ = y (x ≠ 0)
Solution
y = Ax 
Differentiating both sides with respect to x, we get : 
y' = d/dx (Ax) 
⇒ y' = A 
Substituting the value of y' in the given differential equation we get : 
L.H.S. = xy' = x.A = Ax = y = R.H.S
Hence, the given function is the solution of the corresponding differential equation.

6. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation y = x sin x : xy′ = y + x√(x² - y² ) (x ≠ 0 and x > y, or x < -y) 
Solution
y = x sin x
Differentiating both sides of this equation with respect to x, we get : 

7. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation xy = log y + C : y' = y²/(1 - xy)(xy ≠ 1) 
Solution
xy = log y + C 
Differentiating both sides of this equation with respect to x, we get : 

8. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
y – cos y = x : (y sin y + cos y + x) y′ = y
Solution
y - cos y = x ...(1) 
Differentiating both sides of the equation with respect to x, we get : 

⇒ y' + sin y . y' = 1 
⇒ y'(1 + sin y( = 1 
⇒ y' = 1/(1 + sin y) 
Substituting the value of y' in equation (1), we get : 
L.H.S = (y sin y + cos y + x)y' 
(y sin y + cos y + y - cos y) × ([1/(1 + siny)] 
= y(1 + sin y). 1/(1 + sin y) 
= y 
= R.H.S. 
Hence, the given function is the solution of the corresponding differential equation.

9. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation
x + y = tan^–1y : y² y′ + y² + 1 = 0
Solution
x + y = tan^-1 y 
Differentiating both sides of this equation with respect to x, we get: 

= -1 - y² + y² + 1 
= 0 
= R.H.S. 
Hence, the given function is the solution of the corresponding differential equation.

10. verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation
y = √(a² - x² x) ∈ (-a, a) : x + y (dy/dx) = 0 (y ≠ 0) 
Solution
y = √(a² - x² )
Differentiating both sides of this equation with respect to x, we get : 

= x - x
= 0
= R.H.S  
Hence, the given function is the solution of the corresponding differential equation.

11. The numbers of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
Solution
We know that the number of constants in the general solution of a differential equation of order n is equal to its order.
Therefore, the number of constants in the general equation of fourth order differential equation is four.
Hence, the correct answer is D.

12.The numbers of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B) 2
(C) 1
(D) 0
Solution
In a particular solution of a differential equation, there are no arbitrary constants. 
Hence, the correct answer is D.

Exercise 9.3

1. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
x/a + y/b = 1 
Solution
x/a + y/b = 1
Differentiating both sides of the given equation with respect to x, we get : 
1/a + (1/b)(dy/dx) = 0
⇒ 1/a + (1/b)y' = 0
Again , differentiating both sides with respect to x, we get : 
0 + (1/b)y" = 0
⇒ (1/b)y" = 0
⇒ y" = 0
Hence, the required differential equation of the given curve is y" = 0.

2. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y² = a (b² – x²)
Solution
y² = a (b² – x²)
Differentiating both sides with respect to x, we get: 

⇒ 2yy' = -2ax 
⇒ yy' = -ax                ...(1)
Again, differentiating both sides with respect to x, we get :
y'. y' + yy" = -a 
⇒ (y')² + yy" = -a                ...(2) 
Dividing equation (2) by equation (1), we get :

3. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y = a e^3x + b e^–2x. 
Solution
y' = 3ae^3x - 2be^-2x ...(2)
Again, differentiating both sides with respect to x, we get : 

4. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y = e^2x (a + bx)
Solution
y = e^2x (a + bx)
Differentiating both sides with respect to x, we get : 
y' = 2e^2x (a + bx) + e^2x.b 
⇒ y' = e^2x (2a + 2bx + b)  ...(2) 
Multiplying equation (1) with 2 and then subtracting it from equation (2),we get : 

5. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y = e^x (a cos x + b sin x)
Solution
y = e^x (a cos x + b sin x) 
Differentiating both sides with respect to x, we get :  
y' = e^x (a cos x + b sin x) + e^x (- a sin x + b cos x)
= y' = e^x [(a + b) cos x - (a - b) sin x] ...(2) 
Again, differentiating with respect to x, we get : 

⇒ 2y + y" = 2y' 
⇒ y" - 2y' + 2y = 0 
This is the required differential equation of the given curve.

6. Form the differential equation of the family of circles touching the y-axis at the origin.
Solution
The centre of the circle touching the y - axis at origin lies on the x - axis. 
Let (a, 0) be the centre of the circle. 
Since it touches the y - axis at origin, its radius is a. 
now, the equation of the circle with centre (a, 0) and radius (a) is 

7. Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solution
The equation of the parabola having the vertex at origin and the axis along the positive y axis is : 
x² = 4ay                ...(1)

8. Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.
Solution
The equation of the family of ellipses having foci on the y - axis and the centre at origin is as follows: 
x² /b²  + y² /a²  = 1 ...(1) 

9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.
Solution
The equation of the family of hyperbolas with the centre at origin and foci along the x - axis is : 
x² /a²  + y² /b²  = 1 ...(1) 

10. Form the differential equation of the family of circles having centre on y-axis and radius 3 units.
Solution
Let the centre of the circle on y - axis be (0, b). 
The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows : 
x² + (y - b)² = 3² 
⇒ x² + (y - b)² = 9 ...(1) 

Differentiating equation (1) with respect to x, we get : 
2x + 2(y - b). y' = 0 
⇒ (y - b). y' = -x 
⇒ y - b = -x/y' 
Substituting the value of (y - b) in equation (1), we get : 

11. Which of the following differential equations has y = c₁ e^x + c₂ e^–x as the general solution ?

Solution 

The given equation is : 
y = c₁ e^x + c₂ e^–x ...(1)
Differentiating with respect to x, we get : 
dy/dx = c₁e^x - c₂e^–x    
Again, differentiating with respect to x, we get : 

This is the required differential equation of the given equation of curve. 
Hence, the correct answer is B.

12. Which of the following differential equation has y = x as one of its particular solution ? 

Solution
The given equation of curve is y = x. 
Differentiating with respect to x, we get :  
dy/dx =1 ...(1)
Again, differentiating with respect to x, we get : 
d² y/dx² = 0 ...(2) 
Now, on substituting the values of y, d² y/dx², and dy/dx from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct. 

Exercise 9.4

1. For the differential equations find the general solution :  
dy/dx = (1 - cos x)/(1 + cos x) 
Solution
The given differential equation is :  

2. For the differential equation find the general solution : 
dy/dx = √(4 – y²) (-2 < y < 2) 
Solution
The given differential equation is : 
dy/dx = √(4 – y²) 
Separating the variables, we get :  
⇒ dy/√(4 – y²)  = dx 
Now, integrating both sides of this equation, we get :  

⇒ y/2 = sin(x + C) 
⇒ y = 2 sin (x + C) 
This is the required general solution of the given differential equation.

3. For the differential equations find the general solution : 
dy/dx + y = 1(y ≠ 1) 
Solution
The given differential equation is : 
dy/dx + y = 1 
⇒ dy + y dx = dx 
⇒ dy = (1 - y) dx
Separating the variables, we get : 
⇒ dy/(1 - y) = dx 
Now, integrating both sides, we get : 

⇒ y = 1 + Ae^-x (where A = -1/C) 
This is the required general solution of the given differential equation.

4. For the differential equations find the general solution:
sec² x tan y dx + sec² y tan x dy = 0
Solution
The given differential equation is : 
sec² x tan y dx + sec² y tan x dy = 0  

Substituting these values in equation (1), we get :  
log (tan x) = - log (tan y) + log C 
⇒ log (tan x) = log (C/tan y) 
⇒ tan x = C/tan y 
⇒ tan x tan y = C 
This is the required general solution of the given differential equation.

5. For the differential equations find the general solution:
(e^x + e^–x) dy – (e^x – e–x) dx = 0
Solution
The given differential equation is :
(e^x + e^–x) dy – (e^x – e^-x ) dx = 0
⇒ (e^x + e^-x ) dy = (e^x - e^-x )dx 

⇒ (e^x + e^–x) dx = dt 
Substituting his value in equation (1), we get : 
y = ∫(1 + t) dt  + C
⇒ y = log(t) + C 
⇒ y = log (e^x + e^–x) + C
This is the required general solution of the given differential equation.

6. For the differential equations find the general solution: 
dy/dx = (1 + x² )(1+ y²) 
Solution
The given differential equation is :  
dy/dx = (1 + x² )(1+ y²)  
⇒ dy/(1 + y² ) = (1 + x²)dx 
Integrating both sides of this equation, we get :  

⇒ tan^-1 y = ∫dx + ∫x² dx 
⇒ tan^-1 y = x + x³/3 + C 
This is the required general solution of the given differential equation.

7. For the differential equations find the general solution :  
y log y dx - x dy = 0 
Solution
The given differential equation is : 
y log y dx - x dy = 0 
⇒ y log y dx = x dy 
⇒ dy/(y log y) = dx/ x  
Integrating both sides, we get : 
∫dy/(y log y) = ∫ dx/x  ...(1)
Let log y = t 

Substituting this value in equation (1), we get :  
 ∫dt/t =  ∫dx/x 
⇒ log t = log x + log C
⇒ log (log y) = log Cx
⇒ log y = Cx
⇒ y = e^cx
This is the required general solution of the given differential equation.

8. For the differential equations find the general solution :  
x⁵ (dy/dx) = -y⁵ 
Solution
The given differential equation is : 

⇒ x^-4 + y^-4 = -4k 
⇒ x^-4 + y^-4 = C (C = -4k) 
This is the required general solution of the given differential equation.

9. For the differential equation find the general solution : 
e^x tan y dx + (1 – e^x) sec² y dy = 0 
Solution
The given differential equation is : 
dy/dx = sin^-1 x 
⇒ dy = sin^-1 x dx 
Integrating both sides, we get : 

⇒ y = x sin^-1 x + √t + C 
⇒ y = x sin^-1 x + √(1 - x²) + C
This is the required general solution of the given differential equation.

10. For the differential equations find the general solution:
e^x tan y dx + (1 – e^x) sec² y dy = 0 
Solution
The given differential equation is :  
e^x tan y dx + (1 – e^x) sec² y dy = 0 
(1 - e^x )sec² y dy = -e^x tan y dx 
Separating the variables, we get : 

⇒ log (tan y) = log (1 - e^x ) + log C 
⇒ log (tan y) = log [C(1 - e^x )] 
⇒ tan y = C(1 - e^x )
This is the required general solution of the given differential equation.

11. For each of the differential equations find a particular solution satisfying the given condition:
(x³ + x² + x + 1) dy/dx = 2x² + x; y = 1 when x = 0
Solution
The given differential equation is :  

⇒ 2x² + x = Ax² + A + Bx² + Bx + Cx + C 
⇒ 2x² + x = (A + B)x² + (B + C)x + (A + C) 
Comparing the coefficients of x² and x, we get : 
A +  B = 2
B + C = 1
A + C = 0
Solving these equations, we get :  
A = 1/2, B = 3/2 and C = -1/2 
Substituting the values of A, B and C in equation (2), we get :  

Now, y = 1 when x = 0. 
⇒ 1 = (1/4) log (1) - (1/2) tan^-1 0 + C 
⇒ 1 = (1/4) × 0 - (1/2) × 0 + C 
⇒ C = 1 
Substituting C = 1 in equation (3), we get : 
y = [(1/4) log (x + 1)² (x² + 1)³] - (1/2) tan^-1 x + 1

12. For each of the differential equations find a particular solution satisfying the given condition : 
x(x² - 1) dy/dx = 1, y = 0 when x = 2 
Solution
x(x² - 1) dy/dx = 1

Comparing the coefficients of x², x and constant, we get : 
A = -1
B - C = 0
A + B + C = 0
Solving these equations, we get B = 1/2 and C = 1/2 . 
Substituting the values of A, B, and C in equation (2), we get :  

13. For each of the differential equations find a particular solution satisfying the given condition:
cos (dx/dy) = a (a ∈ R); y = 1 when x = 0 
Solution
cos (dy/dx) = a 
⇒ dy/dx = cos^-1 a 
⇒ dy = cos^-1 a dx 
Integrating both sides, we get : 
∫dy = cos^-1 a ∫dx 
⇒ y = cos^-1 a . x + C 
⇒ y = x cos^-1 a + C  ...(1) 
Now, y = 1 when x = 0 
⇒ 1 = 0. cos^-1 a + C 
⇒ C = 1 
Substituting C = 1 in equation (1), we get :  
y = x cos^-1 a + 1 
⇒ (y -1)/x = cos^-1 a  
⇒ cos[(y -1)/x] = a 

14. For each of the differential equations find a particular solution satisfying the given condition:
dy/dx = y tan x; y = 1 when x = 0
Solution 
dy/dx = y tan x 
⇒ dy/y = tan x dx 
Integrating both sides, we get :  
∫dy/y = ∫tan x dx 
⇒ log y = log(sec x) + log C 
⇒ log y = log (C sec x) 
⇒ y = C sec x  ...(1) 
Now, y = 1 when x = 0. 
⇒1 = C × sec 0 
⇒1 = C × 1
⇒ C = 1 
Substituting C = 1 in equation (1), we get :  
y = sec x 

15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e^x sin x.
Solution
The differential equation of the curve is :  
y' = e^x sin x
⇒ dy/dx = e^x sin x
⇒ dy = e^x sin x dx 
Integrating both sides, we get :  
∫dy = ∫e^x sin x dx ...(1) 
Let I = ∫e^x sin x dx.

⇒ I = e^x sin x - e^x sin x - I 
⇒ 2I = e^x (sin x - cos x) 
⇒ I = [e^x (sin x - cos x)]/2 
Substituting this value in equation (1), we get : 
y = [e^x (sin x - cos x)]/2 + C ...(2) 
Now, the curve passes through point (0, 0) 

⇒ 2y = e^x (sin x - cos x) + 1 
⇒ 2y - 1 = e^x (sin x - cos x) 
Hence, the required equation of the curve is 2y - 1 = e^x (sin x - cos x).

16. For the differential equation xy (dy/dx) = (x + 2)(y + 2) find the solution curve passing through the point (1, -1).
Solution
The differential equation of the given curve is :  

⇒ y - 2log(y + 2) = x + 2log x + C 
⇒ y - x - C = log x² + log (y + 2)² 
⇒ y - x - C = log [x² (y + 2)² ]  ...(1) 
Now, the curve passes through point (1, -1). 
⇒ -1 - 1 - C = log[(1)² (-1 + 2)² ] 
⇒ -2 - C = log  1 = 0 
⇒ C = -2
Substituting C = -2 in equation (1), we get :  
y - x + 2 = log[x² (y + 2)² ]
This is the required solution of the given curve.

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
Let x and y be the x - coordinate and y - coordinate of the point on the curve respectively. 
We know that the slope of a tangent to the curve in the coordinate axes is given by the relation. 
dy/dx 
According to the given information, we get :  
y (dy/dx) = x 
⇒y dy = x dx  
Integrating both sides, we get:  
∫y dy = ∫x dx 
⇒ y²/2 = x² /2 + C 
⇒ y² - x² = 2C ...(1) 
Now, the curve passes through point (0, -2). 
∴ (-2)² - 2² = 2C 
⇒ 2C = 4 
Substituting 2C = 4 in equation (1), we get : 
y² - x² = 4 
This is the required equation of the curve.

18. At any point (x, y) 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
It is given that (x, y) is the point of contact of the curve and its tangent. 
The slope (m₁ ) of the line segment joining (x, y) and (-4, -3) is (y +3)/(x + 4) 
We know that the slope of the tangent to the curve is given by the relation, 
dy/dx 
∴ Slope (m₂ ) of the tangent = dy/dx 
According to the given information :  
m₂ = 2m₁ 

⇒ log (y + 3) = 2 log (x + 4) + log C 
⇒ log (y + 3) = log C (x + 4)² 
⇒ y + 3 = C(x + 4)² ...(1) 
This is the general equation of the curve. 
It is given that it passes through point (-2, 1). 
⇒ 1 + 3 = C(-2 + 4)²   
⇒ 4 = 4C 
⇒ C = 1 
Substituting C = 1 in equation (1), we get : 
y + 3 = (x + 4)²  
This is the required equation of the curve. 

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
Let the rate of change of the volume of the balloon be k (where k is a constant ) 

⇒ 4πr³ = 3(kt + C)  ...(1) 
Now, at  t = 0, r = 3,
⇒ 4π × 3³ = 3 (k × 0 + C)
⇒ 108π = 3C
⇒ C = 36π
At t = 3, r = 6:
⇒ 4π × 6³ = 3 (k × 3 + C)
⇒ 864π = 3 (3k + 36π)
⇒ 3k = –288π – 36π = 252π
⇒ k = 84π
Substituting the values of k and C in equation (1), we get: 
4πr³ = 3(84πt + 36π)
⇒ 4πr³ = 4π(63t + 27)
⇒ r³ = 63t + 27 
⇒ r = (63t + 27)^1/3 
Thus, the radius of the balloon after t seconds is (63t + 27)^1/3 . 

20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (log­_e2 = 0.6931).
Solution
Let p, t,  and r represent the principal, time, and rate of interest respectively. 
It is given that the principal increases continuously at the rate of r% per year. 

⇒ r/10 = 0.6931 
⇒ r = 6.931 
Hence, the value of r is 6.93 %.

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 (e^0.5 = 1.648).
Solution
Let p and t be the principal and time respectively. 
It is given that the principal increases continuously at the rate of 5 % per year. 

Now, when t = 0, p = 1000. 
1000 = e^c  ...(2)
At t = 10, years the amount will worth Rs 1648. 

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 y be the number of bacteria at any instant t. 
It is given that the rate of growth of the bacteria is proportional to the number peresent. 
∴ dy/dt  ∝ y 
⇒ dy/dt = ky  (where k is a constant) 
⇒ dy/y = kdt 
Integrating both sides, we get:  
∫dy/y = k ∫dt 
⇒ log y = kt + C ...(1) 
Let y₀ be the number of bacteria at t = 0 .
log y₀ be the number of bacteria at t = 0. 
log y₀ = C 
Substituting the value of C in equation (1), we get : 
log y = kt + log y₀ 
⇒ log y - log y₀ = kt 
⇒ log (y/y₀) = kt 
⇒ kt = log (y/y₀) ...(2) 
Also, it is given that the number of bacteria increases by 10% in 2 hours.  
⇒ y = (110/100)y₀ 
⇒ y/y₀ = 11/10 ...(3) 
Substituting this value in equation (2), we get :  
k⋅2 = log (11/10) 
⇒ k = (1/2) log (11/10)
Therefore, equation (2) becomes : 

23. The general solution of the differential equation dy/dx = e^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

Integrating both sides, we get :  
∫e^–y dy = ∫e^x dx 
⇒ -e^–y = e^x + k 
⇒ e^x + e^–y = -k 
⇒ e^x + e^–y = c     (c = -k ) 
Hence, the correct answer is A.

Exercise 9.5

1. Show that the given differential equation is homogeneous and solve each of them
(x² + xy) dy = (x² + y²) dx
Solution
The given differential equation i.e. (x² + xy) dy = (x² + y²) dx can be written as :

2. Show that the given differential equation is homogeneous and solve each of them  
y' = (x + y)/x 
Solution
The given differential equation is : 

Thus, the given equation is a homogeneous equation . 
To solve it, we make the substitution as : 
y = vx 
Differentiating both sides with respect to x, we get : 
dy/dx = V + x(dv/dx) 
Substituting the values of y and dy/dx in equation (1), we  get :  

Integrating both sides, we get :  
v = log x + C 
⇒ y/x = log x + C 
⇒ y = x log x + Cx 
This is the required solution of the given differential equation.

3. Show that the given differential equation is homogeneous and solve each of them
(x – y) dy – (x + y) dx = 0
Solution
The given differential equation is :  
(x - y)dy - (x + y)dx = 0 

4. Show that the given differential equation is homogeneous and solve each of them
(x² – y²) dx + 2xy dy = 0 
Solution
The given differential equation is : 

5. Show that the given differential equation is homogeneous and solve each of them
x² (dy/dx) = x² - 2y² + xy 
Solution
The given differential equation is : 

6. Show that the given differential equation is homogeneous and solve each of them
x dy -y dx = √(x² + y²)dx
Solution

7. Show that the given differential equation is homogeneous and solve each of them
{x cos (y/x) + y sin (y/x)} ydx = {y sin (y/x) - x cos (y/x)} x dy 
Solution 
The given differential equation is  : 

Therefore, the given differential equation is a homogeneous equation. 
To solve it, we make the substitution as : 
y = vx 
⇒ dy/dx = v + x = dv/dx 
Substituting the values of y and dy/dx in equation (1), we get : 

8. Show that the given differential equation is homogeneous and solve 
x (dy/dx) - y + x sin (y/x) = 0 
Solution

9. Show that the given differential equation is homogeneous and solve 
y dx + x log (y/x)dy - 2xdy = 0 
Solution
y dx + x log(y/x) dy - 2xdy = 0 

10. Show that the given differential equation is homogeneous and solve 
(1 + e^x/y ) dx + e^x/y (1 - x/y)dy = 0 
Solution

11. For the differential equations find the particular solution satisfying the given condition:
(x + y) dy + (x – y) dx = 0; y = 1 when x = 1
Solution
(x + y)dy + (x - y)dx = 0 
⇒ (x + y)dy = -(x - y) dx 

Now, y = 1 at x = 1. 
⇒ log 2 + 2 tan^-11 = 2k 
⇒ log 2 + 2 × π/4 = 2k 
⇒ π/2 + log 2 = 2k 
Substituting the value of 2k in equation (2), we get : 
log(x² + y²) + 2 tan^-1(y/x) = π/2 + log 2 
This is the required solution of the given differential equation. 

12.For the differential equations find the particular solution satisfying the given condition:
x² dy + (xy + y²) dx = 0; y = 1 when x = 1
Solution
 x² dy + (xy + y² )dx = 0 
⇒ x² dy = -(xy + y² )dx 

Now, y =1 at x = 1. 
⇒ 1/(1+2) = C² 
⇒ C² = 1/3 
Substituting C² = 1/3 in equation (2), we get:  
x² y/(y + 2x) = 1/3 
⇒ y + 2x = 3x² y
This is the required solution of the given differential equation.

13. For the differential equations find the particular solution satisfying the given condition:
[x sin² (y/x - y)] dx + xdy = 0; y = π/4 when x = 1
Solution

Therefore, the given differential equation is a homogeneous equation. 
To solve this differential equation, we make the substitution as :  

Integrating both sides, we get : 
- cot v = -log |x| - C 
⇒ cot v = log |x| + C 
⇒ cot(y/x) = log |x| + log C 
⇒ cot (y/x) = log |Cx|  ...(2) 
Now, y = π/4 at x = 1. 
⇒ cot (π/4) = log |C| 
⇒ 1 = log C 
⇒ C = e¹ = e 
Substituting C = e in equation (2), we get :  
cot (y/x) = log |ex| 
This is the required solution of the given differential equation . 

14. For the differential equations find the particular solution satisfying the given condition:
dy/dx - y/x + cosec(y/x) = 0; y = 0 when x = 1 
Solution

Integrating both sides, we get :  
cos v = log x + log C = log |Cx| 
⇒ cos (y/x) = log |Cx| ...(2) 
This is the required solution of the given differential equation. 
Now, y = 0 at x = 1. 
⇒ cos (0) = log C 
⇒ 1 = log C 
⇒ C = e¹ = e 
Substituting C = e in equation (2), we get : 
cos(y/x) = log |(ex)|
This is the required solution of the given differential equation. 

15. For the differential equations find the particular solution satisfying the given condition:
2xy + y² - 2x² (dy/dx) = 0; y = 2 when x = 1 
Solution

16. A homogeneous differential equation of the from dx/dt = h(x/y) can be solved by making the substitution 
(A) y = vx 
(B) v = yx 
(C) x = vy 
(D) x = v
Solution
For solving the homogeneous equation of the form dx/dy = h(x/y), we need to make the substitution as x = vy. 
Hence, the correct answer is C. 

17. Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
(B) (xy) dx – (x³ + y³) dy = 0
(C) (x³ + 2y²) dx + 2xy dy = 0
(D) y² dx + (x² – xy – y²) dy = 0 
Solution
Function F(x, y) is said to be the homogenous function of degree n, if
F(λx, λy) = λⁿ F(x, y) for any non - zero constant (λ). 
Consider the equation given in alternative D : 
y² dx + (x² - xy - y² )dy = 0 

Exercise 9.6

1. For the differential equation find the general solution :  
dy/dx + 2y = sin x 
Solution
The given differential equation is dy/dx + 2y = sin x
This is in the form of dy/dx + py = Q (where p = 2 and Q = sin x)
Now, I.F = e^∫pdx = e^∫3dx  = e^3x
The solution of the given differential equation is given by the relation, 
y(I.F.) = ∫(Q × I.F.) dx  + C
⇒ ye^2x = ∫sin x. e^2x dx + C ...(1)
Let I = ∫sin x. e^2x

Therefore, equation (1) becomes:
ye^2x = e^2x/5 (2 sin x - cos x) + C
⇒ y = 1/5 (2 sin x - cos x) + Ce^-2x
This is the required general solution of the given differential equation.

2. For the differential equations find the general solution:  
dy/dx + 3y = e^-2x 
Solution
The given differential equation is dy/dx + py = Q (where p = 3 and Q = e^-2x)
Now, I.F = e^∫pdx = e^∫3dx = e^3x
The solution of the given differential equation is given by the relation,

3. For the differential equations find the general solution :
dy/dx + y/x = x²
Solution
The given differential equation is:
dy/dx + py = Q (where p = 1/x and Q = x²)

4. For the differential equations find the general solution :  
dy/dx + sec xy = tan x(0 ≤ x < π/2)
Solution
The given differential equation is : 
dy/dx + py = Q  (where p = sec x and Q = tan x) 
Now, I.F = e^∫pdx = e^{∫secx dx} = e^{log(secx + tanx)} = sec x + tan x. 
The general solution of the given differential equation is given by the relation, 
y = (I.F.) = ∫(Q × I.F.) dx + C 
⇒ y(sec x + tan x) = ∫tan x(sec x + tan x)dx + C 
⇒ y(sec x + tan x) = ∫sec x tan x dx + ∫tan² x dx + C 
⇒ y(sec x + tan x) = sec x + ∫(sec² x - 1)dx + C 
⇒ y(sec x + tan x) = sec x + tan x - x + C

5. For the differential equations find the general solution : 
cos² x (dy/dx) + y = tan x(0 ≤ x < π/2)
Solution
The given differential equation is : 

6. For the differential equations find the general solution: 
x(dy/dx) + 2y = x² log x
Solution
The given differential equation is:

7. For the differential equations find the general solution: 
x log x(dy/dx) + y = (2/x) log x
Solution
The given differential equation is :  

Substituting the value of ∫[(2/x²)log x] dx in equation (1), we get :
y log x = (-2/x) (1 + log x) + C
This is the required general solution of the given differential equation.

8. For the differential equations find the general solution: (1 + x²) dy + 2xy dx = cot x dx (x ≠ 0). 
Solution
(1 + x² )dy + 2xy dx = cot x dx 

9. For the differential equations find the general solution: 
x(dy/dx) + y - x + xy cot x = 0 (x ≠ 0) 
Solution

10. For the differential equations find the general solution : 
(x + y) dy/dx = 1 
Solution
(x + y) dy/dx  1 
⇒ dy/dx = 1/(x + y) 
⇒ dx/dy = x + y
⇒ (dx/dy) - x = y
This is a linear differential equation of the form : 
dy/dx + px = Q (where p = -1 and Q = y) 

11. For the differential equations find the general solution: y dx + (x – y²) dy = 0
Solution
y dx + (x - y²)dy = 0 
⇒ y dx = (y² - x)dy 

This is a linear differential equation of the form : 
dy/dx + px = Q (where p = 1/y and Q = y) 

The general solution of the given differential equation is given by the relation, 
x(I.F.) = ∫(Q × I.F.) dy + C 
⇒ xy = ∫(y.y) dy + C 
⇒ x = ∫ y² dy + C 
⇒ xy = y³ /3 + C 
⇒ x = y²/3 + C/y 

12. For the differential equations find the general solution :  
(x + 3y² )(dy/dx) = y (y > 0) 
Solution

13. For the differential equations given find a particular solution satisfying the given condition :  dy/dx + 2y tan x = sin x ; y = 0 when x = π/3
Solution
The given differential equation is dy/dx + 2y tan x = sin x. 
This is a linear equation of the form:  
dy/dx + py = Q (where p = 2 tan x and Q = sin x)

The general solution of the given differential equation is given by the relation, 
y(I.F.) =  ∫(Q × I.F.)dx + C 
⇒ y(sec² x) = ∫(sin x. sec² x)dx + C 
⇒ y sec² x = ∫(sec x. tan x)dx + C 
⇒ ysec² x = sec x + C ...(1) 
Now, y = 0 at x = π/3 . 
Therefore, 

⇒ 0 = 2 + C
⇒C = -2
Substituting C = -2 in equation (1), we get : 
y sec² x = sec x - 2 
⇒ y = cos x - 2cos² x 
Hence, the required solution of the given differential equation is y = cos x - 2cos² x

14. For the differential equations given find a particular solution satisfying the given condition :   (1 + x²)(dy/dx) + 2xy = 1/(1 + x²); y = 0 when x = 1 
Solution

⇒ y(1 + x² ) = tan^-1 x + C ...(1) 

Now, y = 0 at x = 1. 
Therefore, 

0 = tan^-1 1 + C 
⇒ C = -π/4 
Substituting C = -π/4 in equation (1), we get : 
y(1 + x²) = tan^-1 x - π/4 
This is the required general solution of the given differential equation.

15. For the differential equations given find a particular solution satisfying the given condition : dy/dx - 3y cot x = sin 2x; y = 2 when x = π/2
Solution
The given differential equation is dy/dx - 3y cot x = sin 2x. 
This is a linear differential equation  of the form : 
dy/dx + py = Q (where p = -3 cot x and Q = sin 2x)

⇒ y = -2sin² x + C sin³ x ...(1) 
Now, y = 2 at x = π/2
Therefore, we get:  
2 = -2 + C 
⇒ C = 4 
Substituting C = 4 in equation (1), we get :  
y = -2sin² x + 4sin³ x
⇒ y = 4sin³ x - 2sin² x 
This is the required particular solution of the given differential equation. 

16. Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
Solution
Let F(x, y) be the curve passing through the origin. 
At point (x, y), the slope of the curve will be dy/dx. 
According to the given information: 
dy/dx = x + y
⇒ dy/dx - y = x  
This is a linear differential equation of the form : 
dy/dx + py = Q (where p = -1 and Q = x) 

Substituting in equation (1), we get : 
ye^-x = -e^-x (x + 1) + C 
⇒ y = -(x + 1) + Ce^x 
⇒ x + y + 1 = Ce^x ...(2) 
The curve passes through the origin . 
Therefore, equation (2) becomes : 
1 = C 
C = 1 
Substituting C = 1 in equation (2), we get :  
x + y + 1 =e^x 
Hence, the required equation of curve passing through the origin is x + y + 1 = e^x

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.
Solution
Let F(x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent to the curve at (x, y) is dy/dx.
According to the given information: 
(dy/dx) + 5 = x + y 
⇒ dy/dx - y = x - 5 
This is a linear differential equation of the form : 
dy/dx + py = Q (where p = -1 and Q = x - 5) 

Therefore, equation (1) becomes : 
ye^-x = (4 - x)e^-x  + C
⇒ y = 4 - x + Ce^x  
⇒ x + y - 4 = Ce^x ...(2)
The curve passes through point (0, 2). 
Therefore, equation (2) becomes : 
0 + 2 - 4 = Ce⁰ 
⇒ -2 = C 
⇒ C = -2 
Substituting C = -2 in equation (2), we get : 
x + y - 4 = -2e^x  
⇒ y = 4 - x - 2e^x  
This is the required equation of the curve. 

18. The integrating Factor of the differential equation (dy/dx) - y = 2x² is 
(A) e^-x 
(B) e^-y 
(C) 1/x 
(D) x 
Solution
The given differential equation is :  

This is a linear differential equation of the form :  
dy/dx + py = Q (where p = -1/x and Q = 2x) 
The integrating factor (I.F.) is given by the relation, 
The
Hence, the correct answer is C. 

19. The integrating factor of the differential equation. 
(1 - y²) dx/dy + yx = ay(-1 << 1) is 
Solution
The given differential equation is : 

Miscellaneous Solutions

1. For differential equations given below, indicate its order and degree (if defined).
(i) (d²y/dx² ) + 5x (dy/dx)² – 6y = log x
(ii) (dy/dx)³  + 4 (dy/dx)² + 7y = sin x
(iii) (d⁴y/dx⁴ ) - sin  (d³y/dx³ ) = 0 
Solution
(i) The differential equation is given as :  

The highest order derivative present in the differential equation is d²y/dx². Thus, its order is two . The highest power raised to d²y/dx² is one. Hence, its degree is one. 
(ii) The differential equation is given as :

The highest order derivative present in the differential equation is dy/dx. Thus, its order is one. The highest power raised to dy/dx is three. Hence, its degree is three. 
(iii) The differential equation is given as : 
(d⁴y/dx⁴) - sin(d³y/dx³) = 0 
The highest order derivative present in the differential equation is d⁴y/dx⁴. Thus, its order is four. 
However, the given differential equation is not a polynomial equation. Hence, its degree is not defined.

2. For given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
(i) y = ae^x + be^-x + x² : x (d²y/dx²) + 2(dy/dx) - xy + x² - 2 = 0 
(ii) y = e^x (a cos x + b sin x) : d² y/dx² - 2(dy/dx) + 2y = 0 
(iii) y = x sin 3x : (d²y/dx²) + 9y - 6cos 3x = 0 
(iv) x² = 2y² log y: (x² + y²) dy/dx - xy = 0 
Solution
(i) xy = ae^x + be^-x + x² 
Differentiating both sides w.r.t. x we get  

(ii) y = e^x (a cos x + b sin x) = ae^x cos x + be^x sin x. 
Differentiating both sides with respect to x, we get :  

= e^x [(2b - 2a - 2b + 2a) cos x] + e^x [(-2a - 2b + 2a + 2b) sin x] 
= 0
Hence, the given function is a solution of the corresponding differential equation. 

(iii) y = x sin3x 
Differentiating both sides with respect to x, we get :  

Substituting the value of d² y/dx² in the L.H.S. of the given differential equation, we get : 
d²y/dx² + 9y - 6 cos 3x
= (6.cos 3x - 9x sin 3x) + 9x sin 3x - 6 cos 3x 
= 0 
Hence, the given function is a solution of the corresponding differential equation. 
(iv) x² = 2y² log y 
Differentiating both sides with respect to x, we get : 

= xy - xy 
= 0 
Hence, the given function is a solution of the corresponding differential equation.

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 
⇒ x² + a² - 2ax + 2y² = a² 
⇒ 2y² = 2ax - x² ...(1) 
Differentiating with respect to x, we get :  

Hence, the differential equation of the family of curves is given as dy/dx = (2y² - x²)/4xy.

4. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
Solution 
(x³ - 3xy²)dx = (y³ - 3x² y)dy 
⇒ dy/dx = (x³ - 3xy²)/(y³ - 3x² y) ...(1) 
This is a homogeneous equation. To simplify it, we need to make the substitution as : 
y = vx 

5. 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
The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is :  
(x - a)² + (y - a)² = a² ...(1)

Differentiating equation (1) with respect to x, we get : 
2(x - a) + 2(y - a) dy/dx = 0 
⇒(x - a)+ (y - a)y' = 0 
⇒ x - a + yy' - ay' = 0 
⇒ x + y y' - a(1 + y') = 0 
⇒ a = (x + yy')/(1 + y') 
Substituting the value of a in equation (1), we get : 

Hence, the required differential equation of the family of circles is 
(x - y)² [1 +(y')² ] = (x + yy')².

6. Find the general solution of the differential equation dy/dx + √(1 - y²)/(1 - x²) = 0 
Solution

Integrating both sides, we get : 
sin^-1y = -sin^-1x + C
⇒ sin^-1x + sin^-1y =  C

7. Show that the general solution of the differential equation dy/dx + (y² + y + 1)/(x² + x + 1) = 0 is given by (x + y + 1) = A (1 - x - y - 2xy), where A is parameter. 
Solution

The differential equation of the given curve is :  
sin x cos y dx + cos x sin ydy = 0 
⇒ (sin x cos ydx + cos x sin ydy)/(cos x cos y) = 0 
⇒ tan x dx + tan y dy = 0 
Integrating both sides, we get :  
 log (sec x) + log (sec y) =log C 
log (sec x .sec y) = log C 
⇒ sec x .sec y = C ...(1)
The curve passes through point (0, π/4). 
∴ 1 × √2 = C 
⇒ C = √2
On substituting C = √2 in equation (1), we get : 
sec x . sec y = √2 
⇒ sec x . 1/cos y = √2
⇒ cos y = sec x/√2 
Hence, the required equation of the curve is cos y = sec x/√2. 

9. Find the particular solution of the differential equation (1 + e^2x)dy + (1 + y²) ex dx = 0, given that y = 1 when x = 0. 
Solution
(1 + e^2x )dy + (1 +y² )e^x dx = 0 

Now, y = 1 at x = 0. 
Therefore, equation (2) becomes: 
tan^-1 1 + tan^-1 1 = C 
⇒ π/4 + π/4 = C
⇒ C = π/4 
Substituting C = π/4 in equation (2), we get : 
tan^-1 y + tan^-1 (e^x) = π/2
This is the required particular solution of the given differential equation. 

10. Solve the differential equation ye^x/y dx = (xe^x/y + y²)dy (y ≠ 0) 
Solution

From equation (1) and equation (2), we get :  
dz/dy = 1 
⇒ dz = dy 
Integrating both sides, we get :  
z = y + C 
⇒ e^x/y = y + C 

11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)
Solution
(x - y)(dx + dy) = dx - dy 
⇒ (x - y + 1)dy = (1 - x + y)dx 

Integrating both sides, we get :  
t + log |t| = 2x + C 
⇒ (x - y) + log |x - y| = 2x + C 
⇒ log |x - y| = x + y + C ...(3) 
Now, y = -1 at x = 0 . Therefore, equation (3) becomes : 
log 1 = 0 - 1 + C 
C = 1
Substituting C = 1 in equation (3) we get :  
log |x - y| = x + y + 1 
This is the required particular solution of the given differential equation. 

12. Solve the differential equation [(e^-2√x /√x) – y/√x] dx/dy = 1(x ≠ 0) 
Solution

13. Find a particular solution of the differential equation dy/dx + y cot x = 4x cosec x (x ≠ 0), given that y = 0 when x = π/2
Solution
The given differential equation is :  
dy/dx + y cot x = 4 x cosec x 
This equation is a linear differential equation of the form 
dy/dx + py = Q, where p = cot x and Q = 4x cosec x. 

The general solution of the given differential equation is given by, 

y(I.F.) = ∫(Q × I.F.)dx + C 
⇒ y sin x = ∫(4x cosec x . sin x)dx + C 
⇒ y sin x = 4 ∫x dx + C 
⇒ y sin x = 4. x² /2 + C 
⇒ y sin x = 2x² + C  ...(1)
Now, y = 0 at x = π/2
Therefore, equation (1) becomes :  
0 = 2 × (π²/4) +  C 
⇒ C = -(π²/2) 
Substituting C = -(π²/2) in equation (1), we get :  
y sin x = 2x² - (π²/2) 
This is the required particular solution of the given differential equation. 

14. Find a particular solution of the differential equation (x + 1) dy/dx = 2e^-y - 1, given that y = 0 when x = 0 
Solution

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 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?
Solution
Let the population at any instant (t) be y. 
It is given that the rate of increase of population is proportional to the number of inhabitants at any instant . 

Integrating both sides, we get : 
log y + kt + C              ...(1) 
In the year 1999, t = 0 and y = 20000. 
Therefore, we get : 
log 20000 = C             ...(2) 
In the year 2004, t = 5 and y = 25000. 
Therefore, we get :  
Log 25000 = k. 5 + C 
⇒ log 25000 = 5k + log 20000

⇒ y = 20000× 5/4 × 5/4 
⇒ y = 31250 
Hence, the population of the village in 2009 will be 31250. 

16. The general solution of the differential equation (ydx - xdy)/y = 0
(A) xy = C
(B) x = Cy²
(C) y = Cx
(D) y = Cx²
Solution
The given differential equation is :

Integrating both sides, we get : 
log |x| - log |y| = log k 
⇒ log |x/y| = log k 
⇒ x/y = k 
⇒ y = (1/k)x 
⇒ y = Cx where C = 1/x 
Hence, the correct answer is C. 

17. The general solution of a differential equation of the type dx/dy + P₁x = Q is 

Solution
The integrating factor of the give differential equation dx/dy + P₁x = Q₁ is e^∫p₁^dy. 
The general solution of the differential equation is given by, 
x(I.F.) = ∫(Q × I.F.) dy + C 

18. The general solution of the differential equation e^xdy + (y e^x + 2x) dx = 0 is
(A xe^y + x² = C
(B) xe^y + y² = C
(C) ye^x + x² = C
(D) ye^y + x² = C
Solution
The given differential equation is : 
e^x dy + (ye^x + 2x)dx = 0 
⇒ e^x (dy/dx) + ye^x + 2x = 0 
⇒ dy/dx + y = -2xe^-x 
This is a linear differential equation of the form 
dy/dx + Py = Q, where P = 1 and Q = -2xe^-x. 

The general solution of the given differential equation is given by, 
y(I.F.) = ∫(Q × I.F.) dx + C 
⇒ ye^x  = ∫(-2xe^-x .e^x) dx + C 
⇒ ye^x = -∫2xdx + C 
⇒ ye^x = -x² + C
⇒ ye^x + x² = C
Hence, the correct answer is C.