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Tamilnadu Samacheer Kalvi 12th Maths Book Solutions Chapter 10 Ordinary Differential Equations Ex 10.5

Question 1.
If F is the constant force generated by the motor of an automobile of mass M, its velocity V is given by M\(\frac{d \mathbf{V}}{d t}\) = F – kV, where A is a constant. Express V in terms of t given that V = 0 when t = 0.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 1
Substituting in (1)
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 2

Question 2.
The velocity v, of a parachute falling vertically satisfies the equation, Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 3 where g and k are constants. If v and x are both initially zero, find v in terms of x.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 343
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 344
Squaring on both sides
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 4

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5

Question 3.
Find the equation of the curve whose slope is \(\frac{y-1}{x^{2}+x}\) and which passes through the point (1, 0).
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 5
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 6
This passes through (1, 0)
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 7

Question 4.
Solve the following differential equations:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 8
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 9

(ii) ydx + (1 + x2) tan-1 x dy = 0
Solution:
ydx + (1 + x2) tan-1 xdy = 0
(1 + x2) tan-1x dy = -y dx
Separating the variables
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 10
log y = – log (tan-1 x) + log c
log y + log (tan-1 x) = log c
log (y tan-1 x) = log c
y tan-1 x = c

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5

(iii) sin \(\frac{d y}{d x}\) = a, y (0) = 1
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 11
Given y(0) = 1
i.e., When x = 0, y = 1
∴ 1 = 0(sin-1 a) + c ⇒ c = 1
∴ The solution is y = xsin-1 a + 1
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 12

(iv) Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 13
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 14

(v) (ey + 1) cos x dx + ey sin x dy = 0
Solution:
(ey + 1) cos x dx + ey sin x dy = 0
ey sin x dy = – (ey + 1) cos x dx
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 15
log (ey + 1) = – log sin x + log c
log [(ey + 1) + log sin x = log c
log (ey +1) sin x] = log c
(ey+ 1) sin x = c

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5

(vi) Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 16
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 17

(vii) Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 18
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 19

(viii) x cos y dy = ex(x log x + 1)dx
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 20

(ix) Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 21
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 211

(x) Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 22
Solution:
\(\frac{d y}{d x}\) = tan2 (x +y)
Let u = x + y
Differentiating with respect to ‘x’
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 23
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 24

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 Additional Problems

Question 1.
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 25
Solution:
The given equation can be written as
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 26
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 27

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5

Question 2.
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 28
Solution:
Put x +y = z. Differentiating with respect to x we get
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 29

Question 3.
Find the cubic polynomial in x which attains its maximum value 4 and minimum value 0 at x = -1 and 1 respectively.
Solution:
Let the cubic polynomial bey = f(x). Since it attains a maximum atx = -1 and a minimum at x = 1.
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 30
Separating the variables we have dy = k (x2 – 1) dx
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 31
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 32
When x = – 1, y = 4 and when x =1,7 = 0
Substituting the equation (1) we have
2k + 3c = 12; – 2k + 3c =0
On solving we have k = 3 and c = 2. Substituting these values in (1) we get the required cubic polynomial y = x3 – 3x + 2.

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5

Question 4.
The normal lines to a given curve at each point (x, y) on the curve pass through the point (2, 0). The curve passes through the point (2, 3). Formulate the differential equation representing the problem and hence find the equation of the curve.
Solution:
Slope of the normal at any point P(x, y) = \(-\frac{d x}{d y}\)
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 33
Since the curve passes through (2, 3)
Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 35

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