Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4

Question 1.
Show that each of the following expressions is a solution of the corresponding given differential equation.
(i) y = 2x² ; xy’ = 2y
(ii) y = aex + be-x; y” – y = 0
Solution:
(i) v = 2x2 …(1)
Differential equation: xy’ = 2y
Differentiate with respect to ‘x’

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 1
On simplifying, 2y = xy’
∴ (1) is solution of the given differential equation.

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4

(ii) y = aex + be-x; y”- y = 0.
y = aex + be-x …(1) Differential equation: y” – y = 0
Differentiate with respect to ‘x’
y’ = aex – be-x
Again differentiate with respect to ‘x’
y” = aex + be-x
y” = y ⇒ y” – y = 0
∴ (1) is the solution of the given differential equation.

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4

Question 2.
Find the value of m so that the function y = em solution of the given differential equation.
(i) y’ + 2y = 0
(ii) y”- 5y’ + 6y = 0
Solution:
(i) y’ + 2y = 0 ……. (1)
Given y = emx ……….(2)
Differentiating equation (2) w.r.t x, we get
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 1
∴ comparing equation (1) & (3),
we get m = -2

(ii) y” – 5y’ + 6y = 0 ……. (1)
Given y = emx ……… (2)
Differentiating equation (2) w.r.t ‘x’, we get
\(\frac { dy }{ dx }\) = ex. m
Again differentiating w.r.t ‘x’, we get
\(\frac { d^2y }{ dx^2 }\) = emx. m²
To find the value of m:
Given y” – 5y’ + 6y = 0 ∵ y = emx
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 2
m² – 5m + 6 = 0
(m – 2) (m – 3) = 0
∴ m = 2, 3

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4

Question 3.
The Slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through (2, 5). Find the equation of the curve.
Solution:
The slope of the tangent to the curve at any
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 3
The equation can be written as
4y dy = dx …….. (1)
Integrating equation (1) on both sides, we get
4∫ ydy = ∫ dx
4(\(\frac { y^2 }{ 2 }\)) = x + c
2y² = x + c …….. (2)
Since the curve passes through at (2, 5), we get
2(5)² = 2 + c
50 = 2 + c
50 – 2 = c
48 = c
∴ Substituting the value of c in equation (2), we get
2y² = x + 48 is the required equation of the curve.

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4

Question 4.
Show that y = e-x + mx + n is a solution of the differential equation ex(\(\frac { d^2y }{ dx^2 }\)) – 1 = 0
Given y = e-x + mx + n …….. (1)
Differentiating equation (1) w.r.t ‘x’, we get
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 4
substituting the value of \(\frac { d^2y }{ dx^2 }\) in the given differential equation, we get
ex(\(\frac { d^2y }{ dx^2 }\)) – 1
= ex(e-x) – 1
= en-1 = 1 – 1 = 0
Hence y = e-x + mx + n is the solution of the given differential equation ex \(\frac { d^2y }{ dx^2 }\) – 1 = 0

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4

Question 5.
Show that y = ax + \(\frac { b }{ x }\), x ≠ 0 is a solution of the differential equation x²yⁿ + xy’ – y = 0.
Solution:
Given
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 5
Again differentiating we get
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 6
x² y” + 2xy’ – xy’ – y = 0
x²y” + xy’ – y = 0
∴ y = ax + \(\frac { b }{ x }\) is a solution of the differential equation x²y” + xy’ – y = 0

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4

Question 6.
Show that y = ae-3x + b, where a and b are arbitrary constants, is a solution of the differential equation \(\frac { d^2y }{ dx^2 }\) + 3 \(\frac { dy }{ dx }\) = 0.
Solution:
Given y = ae-3x + b …… (1)
Differentiating equation (1) w.r.t ‘x’, we get
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 7
Therefore, y = ae-3x + b is a solution of the given differential equation.

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4

Question 7.
Show that the differential equation representing the family of curves y² = 2a(x + a2/3), where a is a positive parameter, is (y² – 2xy\(\frac { dy }{ dx }\))³ = 8(y\(\frac { dy }{ dx }\))5
Solution:
Given y² = 2ax + 2a5/3 ……. (1)
Differentiating equation (1) w.r.t ‘x’ we get
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 8
Hence y² = 2ox + 2 a³ is a solution of the differential equation
(y² – 2xy\(\frac { dy }{ dx }\))³ = 8(y\(\frac { dy }{ dx }\))5

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4

Question 8.
Show that y = a cos bx is a solution of the! differential equation \(\frac { d^2y }{ dx^2 }\) + b² y = 0.
Solution:
Given y = a cos bx ……. (1)
Differentiating equation (1) w.r.t ‘x’, we get
\(\frac { dy }{ dx }\) = a(-sin bx)b
\(\frac { dy }{ dx }\) = -ab sin bx
Again differentiating, we get
Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4 9
Therefore, y = a cos bx is a solution of the differential equation \(\frac { d^2y }{ dx^2 }\) + b² y = 0

Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.4