Exercise 5.9 - Chapter 5 - Differential Calculus - 11th Business Maths Guide Samacheer Kalvi Solutions

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 Text Book Back Questions and Answers

Question 1.
Find y₂ for the following functions:
(i) y = e^3x+2
(ii) y = log x + a^x
(iii) x = a cosθ, y = a sinθ
Solution:
(i) y = e^3x+2

(ii) y = log x + a^x

(iii) x = a cosθ, y = a sinθ

Question 2.
If y = 500e^7x + 600e^-7x, then show that y₂ – 49y = 0.
Solution:
y = 500e^7x + 600e^-7x

(or) y₂ – 49y = 0

Question 3.
If y = 2 + log x, then show that xy₂ + y₁ = 0.
Solution:
y = 2 + log x

Question 4.
If = a cos mx + b sin mx, then show that y₂ + m²y = 0.
Solution:
y = a cos mx + b sin mx

= a(-sin mx) . m + b(cos mx) . m
= -am sin mx + bm cos mx
y₂ = -am(cos mx) . m + bm(-sin mx) . m
= -am² cos mx – bm² sin mx
= -m² [a cos mx + b sin mx]
= -m²y
∴ y₂ + m²y = 0

Question 5.

 then show that (1 + x²) y₂ + xy₁ – m²y = 0
Solution:

Differentiating with respect to x, we get
(1 + x²) . 2(y₁) (y₂) + (y₁)² (2x) = 2m²yy₁
Dividing both sides by 2y₁ we get,
(1 + x²) y₂ + xy₁ = m²y
⇒ (1 + x²) y₂ + xy₁ – m²y = 0

Question 6.
If y = sin(log x), then show that x²y₂ + xy₁ + y = 0.
Solution:
y = sin(log x)

Differentiating both sides with respect to x, we get
xy₂ + y₁(1) = -sin(log x) . 1/x
⇒ x[xy₂ + y₁] = -sin(log x)
⇒ x²y₂ + xy₁ = -y
⇒ x²y₂ + xy₁ + y = 0