**NCERT Solutions for Maths Class 12 Exercise 9.3**

Hello Students. Welcome to **maths-formula.com**. In this post, you will find the complete** ****NCERT Solutions for Maths Class 12 Exercise 9.3**.

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Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

**NCERT Solutions for Maths Class 12 Exercise 9.3** helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

All the schools affiliated with CBSE, follow the NCERT books for all subjects. You can check your syllabus from **NCERT Syllabus for Mathematics Class 12**.

**NCERT Solutions for Maths Class 12 Exercise 9.3** are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

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**NCERT Solutions for Maths Class 12 Exercise 9.1**

**NCERT Solutions for Maths Class 12 Exercise 9.2**

**NCERT Solutions for Maths Class 12 Exercise 9.4**

**NCERT Solutions for Maths Class 12 Exercise 9.5**

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**NCERT Solutions for Maths Class 12 Exercise 9.6**

**NCERT Solutions for Maths Class 12 Exercise 9.3**

**In each of the Exercises 1 to 5, form a differential equation
representing the given family of curves by eliminating arbitrary constants a and b.**

**Maths Class
12 Ex 9.3 Question 1.**

**Solution:**

The given equation of the curve is …(i)

Differentiating equation (i) w.r.t. *x*,
we get

1/*a* + *y*’/*b* = 0 …(ii)

Again differentiating equation (ii) w.r.t. *x*,
we get

*y*’’/b = 0 ⇒ *y*’’
= 0

Hence, the differential equation representing the given family of curve is *y*’’ = 0.

**Maths Class
12 Ex 9.3 Question 2.**

*y*²
= *a* (*b*² – *x*²)

**Solution:**

The given equation of the curve is *y*² = *a* (*b*² – *x*²) … (i)

Differentiating equation (i) w.r.t. *x*,
we get

2*yy*’ = *a*(–2*x*) ⇒ *yy*’
= –*ax* …(ii)

Again differentiating equation (ii) w.r.t. *x*,
we get

*yy*’’ + *y*’.* y*’ = –*a * ⇒ *yy*’’
+ *y*’^{2} = –*a * … (iii)

Eliminating a from equation (ii) and (iii), we get

*yy*’
= (*yy*’’ + *y*’^{2})*x *⇒ *xy y*’’ + *xy*’^{2} – *yy*’ = 0

Hence, the differential equation representing the given family of curve is *xy y*’’ + *xy*’^{2} – *yy*’ = 0.

**Maths Class
12 Ex 9.3 Question 3.**

*y*
= *a e*^{3x} + *b e*^{-2x}

**Solution:**

The given equation of the curve is *y* = *a e*^{3x} + *b e*^{-2x} … (i)

Differentiating equation (i) w.r.t. *x*,
we get

*y*’
= *a e*^{3x} × 3 + *b e*^{-2x }× (–2) … (ii)

Multiplying equation (i) by 3 and subtracting it from equation
(ii), we get

*y*’
– 3*y* = –5*b e*^{-2x } … (iii)

Differentiating equation (iii) w.r.t. *x*, we get

*y*’’
– 3*y*’ = –5*b e*^{-2x }× (–2) … (iv)

Multiplying equation (iii) by 2 and adding it with equation (iv),
we get

*y*’’ – 3*y*’ + 2(*y*’ – 3*y*) = 0 ⇒ *y*’’ – *y*’ – 6*y* = 0

Hence, the differential equation representing the given family of
curve is *y*’’ – *y*’ – 6*y* = 0.

**Maths Class
12 Ex 9.3 Question 4.**

*y*
= *e*^{2x} (*a* + *bx*)

**Solution:**

The given equation of the curve is *y* = *e*^{2x} (*a* + *bx*) … (i)

Differentiating equation (i) w.r.t. *x*, we get

*y*’
= *a e*^{2x} × 2 + *b e*^{2x }+ *bx e*^{2x} × 2

*y*’
= *b e*^{2x }+ 2* e*^{2x} (*a* + *bx*)

*y*’
= *b e*^{2x }+ 2*y * ⇒ *y*’
– 2*y* = *b e*^{2x }* * …
(ii)

Differentiating equation (ii) w.r.t. *x*, we get

*y*’’
– 2*y*’ = 2*b e*^{2x } … (iii)

From equations (ii) and (iii), we get

*y*’’
– 2*y*’ = 2(*y*’ – 2*y*) ⇒ *y*’’ – 4*y*’ + 4*y* = 0

Hence, the differential equation representing the given family of
curve is *y*’’ – 4*y*’ + 4*y* = 0.

**Maths Class
12 Ex 9.3 Question 5.**

*y*
= *e ^{x}* (

*a*cos

*x*+

*b*sin

*x*)

**Solution:**

The given equation of the curve is *y* = *e ^{x}* (

*a*cos

*x*+

*b*sin

*x*) … (i)

Differentiating equation (i) w.r.t. *x*, we get

**Maths Class
12 Ex 9.3 Question 6.**

Form the differential equation of the family of circles touching
the *y*-axis at origin.

**Solution:**

The equation of the circle with centre (*a*, 0) and radius *a*, which
touches *y*-axis at origin is

**Maths Class
12 Ex 9.3 Question 7.**

Form the differential equation of the family of parabolas having
vertex at origin and axis along positive *y*-axis.

**Solution:**

The equation of parabola having vertex at the origin and axis
along positive *y*-axis is

**Maths Class
12 Ex 9.3 Question 8.**

Form the differential equation of family of ellipses having foci on
*y*-axis and centre at origin.

**Solution:**

The equation of family of ellipses having foci at *y*- axis and centre at origin is

**Maths Class
12 Ex 9.3 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 the family of hyperbola is

Differentiating both sides w.r.t, *x*,
we get

**Maths Class 12 Ex 9.3 Question 10.**

Form the differential equation of the family of circles having
centre on *y*-axis and radius 3 units.

**Solution:**

Let centre of the circle be (0, *a*) and radius *r* = 3 units

Equation of the circle is *x*² + (*y* – *a*)²
= 9 … (i)

Differentiating equation (i) w.r.t. *x*,
we get

**Maths Class
12 Ex 9.3 Question 11.**

Which of the following differential equations has *y* = *c*_{1}
*e ^{x}* +

*c*

_{2}

*e*

^{-x}as the general solution?

**Solution:**

(B) The given equation is *y* = *c*_{1} *e ^{x}* +

*c*

_{2}

*e*

^{-x }… (i)

Differentiating equation (i), w.r.t. *x*, we get

*y*’
= *c*_{1} *e ^{x}* –

*c*

_{2}

*e*

^{-x }… (ii)

^{}Differentiating equation (ii), w.r.t. *x*, we get

*y*’’
= *c*_{1} *e ^{x}* +

*c*

_{2}

*e*

^{-x }⇒

*y*’’ =

*y*

⇒ *y*’’
– *y *= 0 or
*d*^{2}*y*/*dx*^{2}
– *y *= 0

Hence, the correct answer is option (B).

**Maths Class
12 Ex 9.3 Question 12.**

Which of the following differential equations has *y* = *x*
as one of its particular solution?

**Solution:**

(C) The given equation is *y *= *x ^{ }*… (i)

Differentiating equation (i), w.r.t. *x*, we get

*y'* = 1* ^{ }*… (ii)

^{}Differentiating equation (ii), w.r.t. *x*, we get

*y''* = 0* ^{ }*… (iii)

Equations (i), (ii) and (iii)
satisfy the differential equation given in (C).

Hence, the correct answer is option (C).

**NCERT Solutions for Maths Class 12 Exercise 9.1**

**NCERT Solutions for Maths Class 12 Exercise 9.2**

**NCERT Solutions for Maths Class 12 Exercise 9.4**

**NCERT Solutions for Maths Class 12 Exercise 9.5**

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**NCERT Solutions for Maths Class 12 Exercise 9.6**