Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.3 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.3

Question 1.

Find the differential equation of the family of (i) all non-vertical lines in a plane

(ii) all non horizontal lines in a plane.

Solution:

(i) The equation of the family of non vertical lines in a plane ax + by = 1, b ≠ 0, a ∈ R

Given equation is ax + by = 1 …….. (1)

Differentiating equation (1) with respect to ‘x’ we get

a + b \(\frac { dy }{ dx }\) = 0

∵ 2 arbitrary constant,

∴ Differentiating twice continuously

Again differentiating above equation with respect to ‘x’, we get

0 + b \(\frac { d^2y }{ dx^2 }\) = 0

\(\frac { d^2y }{ dx^2 }\) = 0 [∵ b ≠ 0]

\(\frac { d^2y }{ dx^2 }\) = 0 is a required differential equation.

(ii) The equation of the family of non horizontal

lines in a,plane ax + by = 1, a ≠ 0 & b ∈ R

Given equation is ax + by = 1 ……… (1)

Differentiating equation (1) with respect to ‘y’, we get

a \(\frac { dx }{ dy }\) + b = 0

∵ 2 arbitrary constant,

∴ Differentiating twice continuously

Again differentiating we get

a \(\frac { d^2x }{ dy^2 }\) = 0 [∵ a ≠ 0]

\(\frac { d^2x }{ dy^2 }\) = 0 is a required differential equation.

Question 2.

Form the differential equation of all straight lines touching the circle x² + y² = r²

Solution:

Given circle equation be x² + y² = r²

Let y = mx + c be all straight lines which towards the given circle x² + y² = r²

The condition for y = mx + c ……. (1)

be a tangent to the circle x² + y² = r²

be c² = r² (1 + m²) ⇒ c = r \(\sqrt { 1+m^2 }\)

Substituting c value in equation (1), we get

y = mx + r \(\sqrt { 1+m^2 }\)

y – mx = r\(\sqrt { 1+m^2 }\) ……. (2)

Differentiating equation (2) w.r.t x, we get

\(\frac { dy }{ dx }\) – m = 0

\(\frac { dy }{ dx }\) = m …….. (3)

substituting equation (3) in equation (2), we get;

Which is a required differential equation.

Question 3.

Find the differential equation of the family of circles passing through the origin and having their centres on the x-axis.

Solution:

Given the circles centre on x-axis & the circle is passing through the origin.

Let it be (r, 0) & its radius r.

Equation of the circle is

(x – a)² + (y – b)² = r²

(x – r)² + (y – 0)² = r²

x² – 2xr + r² + y² = r²

x² – 2xr + y² = r² – r²

x² – 2xr + y² = 0 ……. (1)

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

2x – 2r + 2y \(\frac { dy }{ dx }\) = 0 dx

2x + 2y\(\frac { dy }{ dx }\) = 2r

x + y\(\frac { dy }{ dx }\) = r

Substituting r value in equation (1), we get

Which is a required differential equation.

Question 4.

Find the differential equation of the family of all the parabolas with latus rectum 4a and whose axes are parallel to the x-axis.

Solution:

Given the equation of family of parabolas with latus rectum 4a and axes are parallel to x-axis then

(y – b)² = 4a (x – a), where (a, b) is the vertex of parabola.

y² – 2yb + b² = 4ax – 4a² ……. (1)

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

Differentiating equation (2) with respect to ‘x’, we get

yy”+ y’y’ = by”

yy” + y’2 = by” ……. (3)

Substituting the b value in (3), we get

Which is a required differential equation.

Question 5.

Find the differential equation of the family of parabolas with vertex at (0, -1) and having axis along the y axis.

Solution:

Equation of the family of parabolas with vertex at (0, -1) and having axis along the y-axis is

(x – 0)² = 4a(y + 1)

x² = 4a (y + 1) ……. (1)

x² = 4 ay + 4a

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

2x = 4a y’

\(\frac { 2x }{ y’ }\) = 4a

Substituting 4a value in equation (1), we get

x² = \(\frac { 2x }{ y’ }\)(y + 1)

\(\frac { x^2 }{ x }\) = \(\frac { 2 }{ y’ }\)(y + 1)

x = \(\frac { 2 }{ y’ }\) (y + 1)

xy’ = 2 (y + 1)

xy’ = 2y + 2

xy’ – 2y – 2 = 0 is a required differential equation.

Question 6.

Find the differential equations of the family of all the ellipses having foci on the y-axis and centre at the origin.

Solution:

The equation of the family of ellipses having centre at the origin & foci on the y-axis, is given by \(\frac { x^2 }{ a^2 }\) + \(\frac { y^2 }{ b^2 }\) = 1 ……… (1)

where b > a & a, b are the parameters or a,b are arbitrary constant.

Differentiating equation (1) twice successively, (because we have two arbitrary constant) we get

Equ (3) – (2) we get

is the required differential equation.

Question 7.

Find the differential equation corresponding to the family of curves represented by the equation y = Ae^{8x} + Be ^{-8x}, where A and B are arbitrary constants.

Solution:

Given y = Ae^{8x} + Be^{-8x} …….. (1)

where A & B are aribitrary constants differentiating equation (1) twice successively (because we have two arbitrary constant), we get

\(\frac { dy }{ dx }\) = Ae^{8x} + Be^{-8x} (-8)

\(\frac { dy }{ dx }\) = 8Ae^{8x} – 8Be^{-8x} ……… (2)

\(\frac { d^2y }{ dx^2 }\) = 8Ae^{8x} (8) – Be^{-8x} (-8)

= 64Ae^{8x} + 64Be^{-8x}

= 64[Ae^{8x} + Be^{-8x}] ……….. (3)

Substituting eqn (1) in eqn (3), we get

\(\frac { d^2y }{ dx^2 }\) = 64 y

\(\frac { d^2y }{ dx^2 }\) – 64 y = 0 is the required differential equation.

Question 8.

Find the differential equation of the curve represented by xy = ae^{x} + be^{-x} + x²

Solution:

Given xy = ae^{x} + be^{-x} + x² ……… (1)

where a & b are aribitrary constant,

differentiate equation (1) twice successively,

because we have two arbitray constant.

From (1), we get xy – x² = ae^{x} + be^{-x} …….. (4)

Substituting equation (4) in (3), we get

∴ x \(\frac { d^2y }{ dx^2 }\) + \(\frac { 2dy }{ dx }\) – xy + x² – 2 = 0 is the required differential equation.