Students can download 11th Business Maths Chapter 5 Differential Calculus Ex 5.9 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.9

### Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.9 Text Book Back Questions and Answers

Question 1.

Find y_{2} 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

\(\frac{d x}{d \theta}\) = a(-sinθ) = -a sinθ …….. (i)

\(\frac{d y}{d \theta}\) = a(cosθ)

Question 2.

If y = 500e^{7x} + 600e^{-7x}, then show that y_{2} – 49y = 0.

Solution:

y = 500e^{7x} + 600e^{-7x}

(or) y_{2} – 49y = 0

Question 3.

If y = 2 + log x, then show that xy_{2} + y_{1} = 0.

Solution:

y = 2 + log x

Question 4.

If = a cos mx + b sin mx, then show that y_{2} + m^{2}y = 0.

Solution:

y = a cos mx + b sin mx

y_{1} = a \(\frac{d}{d x}\) (cos mx) + b \(\frac{d}{d x}\) (sin mx)

[∵ \(\frac{d}{d x}\) (sin mx) = cos mx \(\frac{d}{d x}\) (mx) = (cos mx) . m]

= a(-sin mx) . m + b(cos mx) . m

= -am sin mx + bm cos mx

y_{2} = -am(cos mx) . m + bm(-sin mx) . m

= -am^{2} cos mx – bm^{2} sin mx

= -m^{2} [a cos mx + b sin mx]

= -m^{2}y

∴ y_{2} + m^{2}y = 0

Question 5.

If y = \(\left(x+\sqrt{1+x^{2}}\right)^{m}\), then show that (1 + x^{2}) y_{2} + xy_{1} – m^{2}y = 0

Solution:

\(y_{1}=\frac{m y}{\sqrt{1+x^{2}}}\)

Squaring both sides we get,

\(y_{1}^{2}=\frac{m^{2} y^{2}}{\left(1+x^{2}\right)}\)

(1 + x^{2}) (\(y_{1}^{2}\)) = m^{2}y^{2}

Differentiating with respect to x, we get

(1 + x^{2}) . 2(y_{1}) (y_{2}) + (y_{1})^{2} (2x) = 2m^{2}yy_{1}

Dividing both sides by 2y_{1} we get,

(1 + x^{2}) y_{2} + xy_{1} = m^{2}y

⇒ (1 + x^{2}) y_{2} + xy_{1} – m^{2}y = 0

Question 6.

If y = sin(log x), then show that x^{2}y_{2} + xy_{1} + y = 0.

Solution:

y = sin(log x)

y_{1} = cos(log x) \(\frac{d}{d x}\) (log x)

y_{1} = cos(log x) . \(\frac{1}{x}\)

∴ xy_{1} = cos(log x)

Differentiating both sides with respect to x, we get

xy_{2} + y_{1}(1) = -sin(log x) . \(\frac{1}{x}\)

⇒ x[xy_{2} + y_{1}] = -sin(log x)

⇒ x^{2}y_{2} + xy_{1} = -y

⇒ x^{2}y_{2} + xy_{1} + y = 0