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# NCERT Solutions for Class 12 Maths Differential Equations

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Differential Equations Exercise 9.3*.

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**NCERT Solutions for Class 12 Maths Differential Equations Exercise 9.3*** NCERT Solutions for Class 12 Maths* are not only the solutions of Maths exercise but it builds your foundation of other important subjects. Getting knowledge of depth concept of

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**CBSE Class 12**^{th}Maths## NCERT Solutions for Class 12 Maths Differential Equations

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.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 1.**

x/a + y/b = 1

**NCERT Solutions:**

Given equation is

Differentiating (1) with respect to x, we get

…(2)

Differentiating (2) with respect to x, we get

Thus, the differential equation of the given equation is y” = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 2.**

y^{2} = a (b^{2} – x^{2})

**NCERT Solutions:**

Given equation is y^{2} = a (b^{2} – x^{2}) …(1)

Differentiating (1) with respect to x, we get

2yy’ = a(– 2x) ⇒ yy’ = –ax …(2)

Differentiating (2) with respect to x, we get

yy” + y’.y’ = –a …(3)

Eliminating a between (2) and (3), we get

yy’ = (yy” + y’.y’)x ⇒ yy’ = x(y’^{2} + yy”)

Thus, the differential equation of the given equation is yy’ = x(y’^{2} + yy”).

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 3.**

y = a e^{3x} + b e^{– 2x}

**NCERT Solutions:**

Given equation is y = a e^{3x} + b e^{– 2x} …(1)

Differentiating (1) with respect to x, we get

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

From (2) – 3 x (1), we get

y’ – 3y = –5be^{–2x} …(3)

Differentiating (3) with respect to x, we get

y” – 3y’ = –5be^{–2x}(–2) …(4)

From 3x(3) + (4), we get

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

Thus, the differential equation of the given equation is y” – y’ – 6y = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 4.**

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

**NCERT Solutions:**

Given equation is y = e^{2x} (a + bx) …(1)

Differentiating (1) with respect to x, we get

Again differentiating with respect to x, we get

Thus, the differential equation of the given equation is y” – 4y’ + 4y = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 5.**

y = e^{x} (a cos x + b sin x)

**NCERT Solutions:**

Given equation is y = e^{x} (a cos x + b sin x) …(1)

⇒ e^{-x}y = (a cos x + b sin x)

Differentiating two times with respect to x, we get

Thus, the differential equation of the given equation is y” – 2y’ + 2y = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 6.**

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

**NCERT Solutions:**

Thus, the differential equation of the family of circles touching the y-axis at origin is 2xyy’ + x^{2} = y^{2}.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 7.**

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

**NCERT Solutions:**

Thus, the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis is xy’ – 2y = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 8.**

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

**NCERT Solutions:**

Thus, the differential equation of the family of ellipses having foci on y-axis and centre at origin is x(y’^{2} + yy”) – yy’ = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 9.**

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

**NCERT Solutions:**

Thus, the differential equation of the family of hyperbolas having foci on x-axis and centre at origin is x(yy” + y’^{2}) – yy’ = 0.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 10.**

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

**NCERT Solutions:**

Thus, the differential equation of the family of circles having centre on y-axis and radius 3 units is x^{2}(y’^{2} + 1) = 9y’^{2}.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 11.**

Which of the following differential equations has y = c_{1} e^{x} + c_{2} e^{–x} as the general solution?

**NCERT Solutions:**

Given equation is y = c_{1} e^{x} + c_{2} e^{–x} …(1)

Differentiating (1) with respect to x, we get

y’ = c_{1} e^{x} – c_{2} e^{–x} …(2)

Differentiating (2) with respect to x, we get

y” = c_{1} e^{x} + c_{2} e^{–x }⇒ y” = y.

Thus, the differential equation of the given equation is y” = y.

**NCERT Solutions for Class 12 Maths Differential Equations ****Exercise**** 9.3: Ques No 12.**

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

**NCERT Solutions:**