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 – c3 ……. (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)
