Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.3

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

Question 1.

\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1

Solution:

\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1

Differentiating w.r.t. x,

\(\frac{1}{a}\) + \(\frac{y’}{b}\) = 0.

Again differentiating,

\(\frac{1}{a}\).y” = 0 or y” = 0.

∴ Required differential equation is y” = 0

Question 2.

y^{2} = a(b^{2} – x^{2})

Solution:

y^{2} = a(b^{2} – x^{2}) ………………….. (1)

Differentiating w.r.t. x,

2yy’ = a(0 – 2x) = – 2ax ………………… (2)

Again differentiating,

2(y^{2} + yy’) = – 2a ………………… (3)

Dividing (3) by (2) we get,

⇒ x(y^{2} + yy”) = yy’,

which is the required differential equation.

It can also be written as xy\(\frac{d^{2} y}{d x^{2}}\) + x(\(\frac{dy}{dx}\))^{2} – y\(\frac{dy}{dx}\) = 0.

Question 3.

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

Solution:

y = a e^{3x} + b e^{-2x} ……………………. (1)

Differentiating w.r.t. x,

y’ = 3ae^{3x} – 2be^{-2x}x ……………….. (2)

Again differentiating,

y” = 9a^{3x} + 4be^{-2}x ………………….. (3)

Multiplying equation (1) by 2 and add with (2) to obtain:

Multiplying (1) by 3 and subtract (2) from it to obtain:

Putting values of ae^{3x} and be^{-2x} in (3), we get

y” = 9(\(\frac{2y+y’}{5}\)) + 4(\(\frac{3y-y’}{5}\))

or 5y” = 9(2y + y’) + 4(3y – y’)

= 30y + 5y’

or y” – y’ – 6y = 0

∴ Required differential equation is

\(\frac{d^{2} y}{d x^{2}}\) – \(\frac{dy}{dx}\) – 6y = 0.

Question 4.

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

Solution:

The curve is y = e^{2x}(a + bx) ………………. (1)

∴ y’ = 2e^{2x}(a + bx) + e^{2x}.

= e^{2x}(2a + b + 2bx) ……………….. (2)

Multiply eq. (1) and (2) and subtract it from (2) to get:

Differentiating eq. (3),

y” – 2y’ = 2be^{2x} …………….. (4)

Dividing (4) by (3), we get

or y” – 2y’ = 2(y’ – 2y)

or y” = 4y’ + 4y = 0

∴ The required differential equation is

\(\frac{d^{2} y}{d x^{2}}\) – 4\(\frac{dy}{dx}\) + 4y = 0.

Question 5.

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

Solution:

The curve is y = e^{x}(a cos x + b sin x) ……………. (1)

Differentiating w.r.t. x, we get

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

= e^{x}[(a + b)cos x – (a – b)sin x] ……………. (2)

Again differentiating, we get

y” = e^{x}(acos x + bsin x) + e^{x}(- asin x + bcos x)

= e^{x}[(a + b)cos x – (a – b)cos x]

= e^{x}[2b cos x – 2a sin x]

= 2e^{x}(b cos x – a sin x)

or \(\frac{y”}{2}\) = e^{x}(b cos x – a sin x) ……………….. (3)

Adding (1) and (3),

y + \(\frac{y”}{2}\) = e^{x}[(a + b)cos x – (a – b)sin x] = y’

or 2y + y” = 2y’

⇒ y” – 2y’ + 2y = 0.

Hence, the required differential equation is

\(\frac{d^{2} y}{d x^{2}}\) – 2\(\frac{dy}{dx}\) + 2y = 0.

Question 6.

From 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

(x – a)^{2} + y^{2} = a^{2}

or x^{2} + y^{2} = 2ax ……………. (1)

Differentiating w.r.t. x,

2x + 2y y’ = 2a

or x + yy’ = a

Putting value of a in (1), we get

x^{2} + y^{2} = 2x(x + yy’)

= 2x^{2} + 2xy y’.

∴ Required differential equation is

2xy\(\frac{dy}{dx}\) + x^{2} – y^{2} = 0.

Question 7.

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

Solution:

Equation of parabola having vertex at the origin and axis along positive y-axis is

x^{2} = 4ay ………….. (1)

where a is the parameter.

Differentiating w.r.t. x,

2x = 4ay’ ………………. (2)

Dividing (2) by (1) we get,

\(\frac{2 x}{x^{2}}\) = \(\frac{2ay’}{4ay}\) ⇒ \(\frac{y’}{y}\) = \(\frac{2}{x}\)

∴ xy’ = 2y.

∴ Required differential equation is

x\(\frac{dy}{dx}\) – 2y = 0.

Question 8.

Form the differential equation of family of ellipses having foci on y-axis.

Solution:

The equation of family of ellipses having foci on y-axis is

\(\frac{x^{2}}{b^{2}}\) + \(\frac{y^{2}}{a^{2}}\) = 1, a > b …………. (1)

Differentiating w.r.t. x,

Again differentiating,

Putting this value of \(\frac{1}{b^{2}}\) in (2), we get

or x(y’^{2} + yy”) = yy’

or xyy” + xy’^{2} – yy’ = 0.

∴ Required differential equation is

xy \(\frac{d^{2} y}{d x^{2}}\) + x(\(\frac{dy}{dx}\))^{2} – y(\(\frac{dy}{dx}\)) = 0.

Question 9.

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

Solution:

Equation on family of hyperbolas with centre at the origin and foci on x-axis is

\(\frac{x^{2}}{a^{2}}\) – \(\frac{y^{2}}{b^{2}}\) = 1 ………………. (1)

Differentiating w.r.t. x,

Again differentiating,

Question 10.

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

Solution:

Let the centre be [0, b].

∴ Equation of family of circles of radius 3 is

x^{2} + (y – b)^{2} = 9 ………….. (1) [∵ r = 3]

Differentiating w.r.t. x,

2x + 2(y – b)y’ = 0 or y – b = – \(\frac{x}{y’}\)

Putting this value of (y – b) in (1), we get

x^{2} + (- \(\frac{x}{y’}\))^{2} = 9

or x^{2}y’^{2} + x^{2} = 9y’^{2} or (x^{2} – 9)y’^{2} + x^{2} = 0

∴ Required differential equation is (x^{2} – 9) (\(\frac{dy}{dx}\))^{2} + x^{2} = 0.

Choose the correct answers in the following questions 11 and 12:

Question 11.

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

(A) \(\frac{d^{2} y}{d x^{2}}\) + y = 0

(B) \(\frac{d^{2} y}{d x^{2}}\) – y = 0

(C) \(\frac{d^{2} y}{d x^{2}}\) + 1 = 0

(D) \(\frac{d^{2} y}{d x^{2}}\) – 1 = 0

Solution:

Family of curve is y = c_{1}e^{x} + c_{2}e^{-x} …………….. (1)

Differentiating w.r.t. x,

y’ = c_{1}e_{x} – c_{2}e^{-x}

y” = c_{1}e^{x} + c_{2}e_{-x} = y

∴ y” – y = 0

Solution is \(\frac{d^{2} y}{d x^{2}}\) – y = 0.

∴ Part (B) is the correct answer.

Question 12.

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

(A) \(\frac{d^{2} y}{d x^{2}}\) – x^{2}\(\frac{dy}{dx}\) + xy = 0

(B) \(\frac{d^{2} y}{d x^{2}}\) + x\(\frac{dy}{dx}\) + xy = x

(C) \(\frac{d^{2} y}{d x^{2}}\) – x^{2}\(\frac{dy}{dx}\) + xy = 0

(D) \(\frac{d^{2} y}{d x^{2}}\) + x \(\frac{dy}{dx}\) + xy = 0

Solution:

The curve is y = x.

Differentiating w.r.t. x,

y’ = 1 and y” = 0.

Now consider part (C).

\(\frac{d^{2} y}{d x^{2}}\) – x^{2}\(\frac{dy}{dx}\) + xy = 0 – x^{2}.1 + x . x

∵ y = x

= – x^{2} + x^{2} = 0.

Hence, Part (C) is the correct answer.