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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
Solution:
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.

(ii) y = aex + be-x ; y” – y = 0
Solution:
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 Solutions Chapter 10 Ordinary Differential Equations Ex 10.4

Question 2.
Find value of m so that the function y = emx is a solution of the given differential equation.
(i) y + 2y = 0
Solution:
Given solution y = emx
Differentiate with respect to ‘x’
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 2

(ii) y” – 5y’ + 6y = 0
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 266
Given differential equation is y” – 5y’ + 6y = 0
Substitute (1), (2) and (3) in this
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 3

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:
Slope of the tangent is the reciprocal of four times the ordinate
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 4

Question 4.
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 5
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 6

Samacheer Kalvi 12th Maths Solutions 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 x2y”+ xy’ – y = 0.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 7
Here ‘a’ and ‘b’ are arbitrary constants
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 8
Differentiate with respect to ‘x’
xy’ + y . 1 = a (2x) = 2ax ……….. (2)
Differentiate again with respect to ‘x’ .
xy” + y’ . 1 + y = 2a ⇒ xy” + 2y’ = 2a …….. (3)
Substitute (3) in (2)
xy’ + y = (xy” + 2y’)x
xy’ + y = x2y” + 2xy’ ⇒ x2y” + xy’ – y = 0
Hence proved.

Question 6.
Show that y = ae-3x + b, where a and b are arbitrary constants, is a solution of the differential equation
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 10
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 11

Question 7.
Show that the differential equation representing the family of curves Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 111 where a is a positive parameter, is Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 15
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 12

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

Question 8.
Show that, y = a cos bx is a solution of the differential equation Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 13.
Solution:
y = a cos bx …(1) (a is an arbitrary constant)
Differentiating with respect to ‘x’
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 13
Again, differentiating with respect to ‘x’
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 14

Hence proved

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

Question 1.
Verify that the function y = a cos x + b sin x is a solution of the differential equation cos \(\frac{d y}{d x}\) + y sin x = b. dx
Solution:
The given function is y = a cos x + b sin x
Differentiating both sides with respect to x, we have
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 155
Putting values of \(\frac{d y}{d x}\) and y in the given differential equation, we have
L.H.S. = cos x (- a sin x + b cos x) + {a cos x + b sin x) sin x
= – a sin x cos x + b cos2 x + a sin x cos x + b sin2 x = b (cos2 x + sin2 x)
= b × 1 = b = R.H.S
Thus, y = a cos x + b sin x is a solution of differential equation
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 166
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4

Question 2.
Verify that the function y = 4 sin 3x is a solution of the differential equation Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 300
Solution:
The given function is y = 4 sin 3x
Differentiating both sides with respect to x, we have
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 301
Again, differentiating both sides with respect to x, we have
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 302
Putting values of \(\frac{d^{2} y}{d x^{2}}\) and y in the given differential equation, we have
L.H.S. = – 36 sin 3x + 9 (4 sin 3x) = – 36 sin 3x + 36 sin 3x = 0 = R.H.S.
Thus, y = 4 sin 3x is a solution of differential equation Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 303

Question 3.
Verify that the function y = ax2 + bx + c is a solution of the differential equation Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 304   .
Solution:
The given function is y = ax2 + bx + c
Differentiating both sides with respect to x, we have
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 305
Again differentiating both sides with respect to x, we have
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 306
Which is the given differential equation.
Thus, y = ax2 + bx + c is a solution of differential equation \(\frac{d^{2} y}{d x^{2}}\)= 2a.

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

Question 4.
Verify that the function y = e-3x is a solution of the differential equation Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 307
Solution:
The given function is y = e-3x
Differentiating both sides with respect to x, we have
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 307
Again, differentiating both sides with respect to x, we have
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 308

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