--Advertisement--

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

Class 12 Maths NCERT Solutions for Chapter 8 Application of Integrals Exercise 9.3

Differential Equations Exercise 9.3 Solutions

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 :

⇒ xyy" + x(y')2 - yy" = 0
This is the required differential equation of the given curve.


3. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y = a e3x + b e2x

Solution

y' = 3ae3x - 2be-2x ...(2)
Again, differentiating both sides with respect to x, we get : 

Multiplying equation (1) with 2 and then adding it to equation (2) , we get : 

Now, multiplying equation (1) with 3 and subtracting equation (2) from it, we get : 

Substituting the values of ae3x and be-2x in equation (3), we get : 

⇒ y'' = 6y + y' 
⇒ y'' - y' - 6y = 0 
This is the required differential equation of the given curve.


4. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y = e2x (a + bx)

Solution

y = e2x (a + bx)
Differentiating both sides with respect to x, we get : 
y' = 2e2x (a + bx) + e2x.b 
⇒ y' = e2x (2a + 2bx + b)  ...(2) 
Multiplying equation (1) with 2 and then subtracting it from equation (2),we get : 

Differentiating both sides with respect to x, we get : 
y" - 2y' = 2be2x ...(4) 
Dividing  equation (4) by equation (3), we get : 

This is the required differential equation of the given curve.


5. Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y = ex (a cos x + b sin x)

Solution

y = ex (a cos x + b sin x) 
Differentiating both sides with respect to x, we get :  
y' = ex (a cos x + b sin x) + ex (- a sin x + b cos x)
= y' = ex [(a + b) cos x - (a - b) sin x] ...(2) 
Again, differentiating with respect to x, we get : 

Adding equations (1) and (3) , 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 

Differentiating equation (1) with respect to x, we get : 
2x + 2yy' = 2a 
⇒ x + yy' = a 
Now, on substituting the value of a in equation (1), we get : 

This is the required differential equation.


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)

Differentiating equation (1) with respect to x, we get : 
2x = 4ay'  ...(2)
Dividing equation (2) by equation (1), we get : 

This is the required differential equation.


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) 

Differentiating equation (1) with respect to x, we get : 

Again, differentiating with respect to x, we get : 

Substituting this value in equation (2), we get : 

This is the required differential equation.


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) 

Differentiating both sides of  equation (1) with respet to x, we get : 

Again, differentiating both sides with respect to x, we get : 

Substituting the value of 1/a2 in equation (2), we get :  

This is the required differential equation.


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 :  
This is the required differential equation. 

11. Which of the following differential equations has y = c1 ex + c2 e–x as the general solution ?
Solution 
The given equation is : 
y = c1 ex + c2 e–x ...(1)
Differentiating with respect to x, we get : 
dy/dx = c1ex - 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. 

= -x2 + x2 
= 0 
Hence, the correct answer is C.

Previous Post Next Post

--Advertisement--

--Advertisement--