Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 4 Differential Equations Ex 4.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.1

Question 1.

Find the order and degree of the following differential equations.

(i) \(\frac { dy }{dx}\) + 2y = x³

Solution:

Highest order derivative is \(\frac { dy }{dx}\)

∴ order = 1

Power of the highest order derivative \(\frac { dy }{dx}\) is 1

∴ degree = 1

(ii) \(\frac { d^3y }{dx^3}\) + 3(\(\frac { dy }{dx}\))³+ 2\(\frac { dy }{dx}\) = 0

Solution:

Highest order derivative is \(\frac { d^3y }{dx^3}\)

∴ order = 3

Power of the highest order derivative \(\frac { d^3y }{dx^3}\) is 1

∴ degree = 1

(iii) \(\frac { d^2y }{dx^2}\) = \(\sqrt{y – \frac { dy }{dx}}\)

Solution:

[ \(\frac { d^2y }{dx^2}\) ]² = y – \(\frac { dy }{dx}\)

Highest order derivative is \(\frac { d^2y }{dx^2}\)

∴ order = 2

Power of the highest order derivative \(\frac { d^2y }{dx^2}\) is 2

∴ degree = 2

(iv) \(\frac { d^3y }{dx^3}\) = 0

Solution:

Highest order derivative is \(\frac { d^3y }{dx^3}\)

∴ order = 3

Power of the highest order derivative \(\frac { d^3y }{dx^3}\) is 1

∴ degree = 1

(v) \(\frac { d^3y }{dx^3}\) + y + [ \(\frac { dy }{dx}\) – \(\frac { d^3y }{dx^3}\) ]^{3/2} = 0

Solution:

Highest order derivative is \(\frac { d^3y }{dx^3}\)

∴ order = 3

Power of the highest order derivative \(\frac { d^3y }{dx^3}\) is 3

∴ degree = 3

(vi) (2 – y”)2 = y”² + 2y’

Solution:

(2)² – 2(2) (y”) + (y”)² = (y”)² + 2y’

4 – 4y” = 2y’

Highest order derivative is y”

∴ order = 2

Power of the highest order derivative y” is 2

∴ degree = 2

(vii) (\(\frac { dy }{dx}\))³ + y = x – \(\frac { dx }{dy}\)

Solution:

Highest order derivative is \(\frac { dy }{dx}\)

∴ order = 1

Power of the highest order derivative \(\frac { dy }{dx}\) is 4

∴ degree = 4

Question 2.

Find the differential equation of the following

(i) y = cx + c – c³

(ii) y = c (x – c)²

(iii) xy = c²

(iv) x² + y² = a²

Solution:

(i) y = cx + c – c^{3} ……. (1)

Here c is a constant which has to be eliminated

Differentiating w.r.t x, \(\frac{d y}{d x}\) = c …… (2)

Using (2) in (1) we get,

\(y=\left(\frac{d y}{d x}\right) x+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^{3}\) which is the required differential equation.

(ii) y = c (x – c)² ……… (1)

y = c (x² – 2cx + c²)

y = cx² – 2c²x + c³

Differentiating w.r. to x

Substituting this value of c and (x – c) in (1), we get

(iii) xy = c²

Differentiating w.r. to x

x(\(\frac { dy }{dx}\)) + y(1) = 0

∴ x(\(\frac { dy }{dx}\)) + y = 0

(iv) x² + y² = a²

Differentiating w.r. to x

Question 3.

Form the differential equation by eliminating α and ß from (x – α)² + (y – α)² = γ²

Solution:

(x – α)² + (y – α)² = γ² ……… (1)

where α and ß are parameters.

Since equation (1) contains two orbitary constants,

We differentiate it two times w.r.t. x

Differentiating (1) w.r.t. x, we get

Substituting the value of (x – α) and (y – ß) in (5) we get

This is the required differential equation.

Question 4.

Find the differential equation of the family of all straight lines passing through the origin.

Solution:

The general equation for a family of lines passing through the origin is

y = mx ……. (1)

Differentiating w.r.t x,

\(\frac{d y}{d x}\) = m ……. (2)

Using (2) in (1)

y = (\(\frac{d y}{d x}\)) x is the required differential equation

Question 5.

Form the differential equation that represents all parabolas each of which has a latus rectum 4a and whose axes are parallel to the x-axis.

Solution:

The equation of the family of the parabola is

(y – k)² = 4a (x – h) ……. (1)

where h and k are arbitrary constants,

[we have to differentiate the equation twice to eliminate h and k]

Differentiating equation (1) w.r.t. x

Question 6.

Find the differential equation of all circles passing through the origin and having their centers on the y axis, x² + (y – k)² = r²

Solution:

Equation of circle whose centre is (h, k)

(x – h)² + (y – k)² = r²

since the centre is on the y-axis (ie) (0, k) be the centre

(x – 0)² + (y – k)² = r²

x² + (y – k)² = r² ………. (1)

since the circle passing the origin (0, 0)

Eqn (1) becomes

0 + (0 – k)² = r²

k² = r² ⇒ r = k

Eqn (1) ⇒ x² + (y – k)² = k²

x² + y² – 2yk + k² = k²

x² + y² – 2yk = 0

x² + y² = 2yk ……….. (2)

Differentiating w.r.t. x

Question 7.

Find the differential equation of the family of parabola with foci at the origin and axis along the x axis, y² = 4a (x + a)

Solution:

Equation of parabola with foci at the origin and axis along the x-axis is

y² = 4a(x + a) ……… (1)

Differentiate w.r.t. x

2y \(\frac { dy }{dx}\) = 4a (1 + 0)

2y = \(\frac { dy }{dx}\) = 4a ⇒ a = \(\frac { y }{2}\), \(\frac { dy }{dx}\)

Substitute the value of a = \(\frac { y }{2}\) \(\frac { dy }{dx}\) in equ (1)