Mathematics-II (BCA-201) - 2018

Section-A

(Very Short Answer Questions)
Each question carries 3 marks. Very short answer is required.

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Question 1:

Define the following with examples:

(i) Proper subset
(ii) Complement of a set
(iii) What is the set {x : x ∈ ℝ, x² = 9, 2x = 4}?

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Solution:

(i) Proper Subset:
A proper subset of a set $A$ is a subset of $A$ that is not equal to $A$.
In other words, if $B \subset A$ and $B \neq A$, then $B$ is a proper subset of $A$.

Example:
Let $A = \{1, 2, 3\}$. Then $B = \{1, 2\}$ is a proper subset of $A$, because $B \subset A$ and $B \neq A$.

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(ii) Complement of a set:
The complement of a set $A$, denoted by $A'$, is the set of all elements in the universal set $U$ that are not in $A$.
Mathematically,
$A' = U - A$

Example:
Let $U = \{1, 2, 3, 4, 5\}$ and $A = \{1, 2, 3\}$. Then the complement of $A$ is:
$A' = \{4, 5\}$

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(iii) Set defined by $\{x : x \in \mathbb{R}, x^2 = 9, 2x = 4\}$:
First, solve the equations:

• From $2x = 4$, divide both sides by 2:
  $x = 2$

• Substitute $x = 2$ into $x^2 = 9$:
  $2^2 = 4 \neq 9$.

Since $x = 2$ does not satisfy $x^2 = 9$, there are no solutions to this set.
Thus, the set is:
$\emptyset$ (empty set).

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Question 2:

Let $f: A \to B$ such that $f(x) = x - 1$ and $g: B \to C$ such that $g(y) = y^2$. Find $g(f(x))$.

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Solution:

We are tasked with finding $g(f(x))$, which is the composition of $g$ and $f$.

1. First, $f(x) = x - 1$.
2. Substitute $f(x)$ into $g(y) = y^2$:
   $g(f(x)) = g(x - 1) = (x - 1)^2$

So,
$g(f(x)) = (x - 1)^2$

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Question 3:

Show that a linearly ordered poset is a distributive lattice.

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Solution:

1. Poset (Partially Ordered Set): A poset is a set $P$ equipped with a partial order $\leq$.

   • A linearly ordered poset means any two elements $a, b \in P$ are comparable, i.e., either $a \leq b$ or $b \leq a$.

2. Lattice: A lattice is a poset where every two elements have a least upper bound (join) and greatest lower bound (meet).

3. Distributive Lattice: A lattice is distributive if, for all $a, b, c \in P$:
   $a \land (b \lor c) = (a \land b) \lor (a \land c)$
   $a \lor (b \land c) = (a \lor b) \land (a \lor c)$

4. Proof:

   • In a linearly ordered poset, the operations $\lor$ (join) and $\land$ (meet) are well-defined because every two elements are comparable.
   • For any $a, b, c \in P$:
     • The join $b \lor c$ is either $b$ or $c$, and the meet $b \land c$ is either $b$ or $c$, based on their order.
   • Substituting into the distributive property equations confirms that they hold.

Thus, a linearly ordered poset is a distributive lattice.

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Question 4:

If $u = \tan^{-1} \left( \frac{y}{x} \right)$, show that $\frac{\partial u}{\partial x} = -\frac{y}{x^2 + y^2}$.

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Solution:

Given:
$u = \tan^{-1} \left( \frac{y}{x} \right)$

1. Differentiate $u$ with respect to $x$, treating $y$ as a constant:
   Using the chain rule:
   $\frac{\partial u}{\partial x} = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \frac{\partial}{\partial x} \left( \frac{y}{x} \right)$

2. Simplify:
   $\frac{\partial}{\partial x} \left( \frac{y}{x} \right) = y \cdot \frac{\partial}{\partial x} \left( x^{-1} \right) = y \cdot (-x^{-2}) = -\frac{y}{x^2}$

3. Substituting back:
   $\frac{\partial u}{\partial x} = \frac{1}{1 + \frac{y^2}{x^2}} \cdot \left( -\frac{y}{x^2} \right)$

4. Combine terms:
   $\frac{\partial u}{\partial x} = \frac{-\frac{y}{x^2}}{\frac{x^2 + y^2}{x^2}}$
   $\frac{\partial u}{\partial x} = -\frac{y}{x^2 + y^2}$

Thus,
$\frac{\partial u}{\partial x} = -\frac{y}{x^2 + y^2}$

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Question 5:

Evaluate $\int_{0}^{\pi} x \sin^2 x \, dx$.

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Solution:

Given:
$I = \int_{0}^{\pi} x \sin^2 x \, dx$

1. Use the trigonometric identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:
   $I = \int_{0}^{\pi} x \cdot \frac{1 - \cos 2x}{2} \, dx$
   $I = \frac{1}{2} \int_{0}^{\pi} x \, dx - \frac{1}{2} \int_{0}^{\pi} x \cos 2x \, dx$

2. Compute the first term:
   $\int_{0}^{\pi} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\pi} = \frac{\pi^2}{2}$

3. Compute the second term using integration by parts:
   Let $u = x$, $dv = \cos 2x \, dx$. Then, $du = dx$, and $v = \frac{\sin 2x}{2}$.
   $\int x \cos 2x \, dx = \frac{x \sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx$
   $= \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} + C$

4. Evaluate:
   Substitute limits $0$ to $\pi$. After simplifications,
   $I = \frac{\pi^2}{4}$

Thus,
$\int_{0}^{\pi} x \sin^2 x \, dx = \frac{\pi^2}{4}$

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Section-B

(Short Answer Questions)
Attempt any two questions out of the following three. Each question carries 7 marks. Short answer is required.

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Question 6:

Show that the direction cosines of a line whose direction ratios are $a, b, c$ are:

$$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \, m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \, n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$$

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Solution:

1. Direction Cosines and Direction Ratios:

   • The direction cosines ($l, m, n$) of a line are the cosines of the angles that the line makes with the coordinate axes ($x, y, z$, respectively).
   • The direction ratios ($a, b, c$) are proportional to the direction cosines.

2. Relationship Between Direction Cosines:
   The direction cosines $l, m, n$ satisfy:

   $$l^2 + m^2 + n^2 = 1$$

3. Normalize the Direction Ratios:
   To compute the direction cosines from the direction ratios $a, b, c$, divide each ratio by the magnitude of the vector formed by $(a, b, c)$:

   $$\text{Magnitude} = \sqrt{a^2 + b^2 + c^2}$$

4. Expressions for Direction Cosines:

   • For the $x$-axis: $$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$$
   • For the $y$-axis: $$m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}$$
   • For the $z$-axis: $$n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$$

Thus, the direction cosines of a line with direction ratios $a, b, c$ are:

$$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \, m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \, n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$$

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Question 7:

In a group of 50 people, 35 speak Hindi, 25 speak both English and Hindi, and all people speak at least one of the two languages. How many people speak only English and not Hindi? How many speak English?

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Solution:

1. Information Given:

   • Total number of people = $50$.
   • Number of people who speak Hindi = $35$.
   • Number of people who speak both English and Hindi = $25$.
   • All people speak at least one language.

2. Let $x$ be the number of people who speak only English.
   From the given data:

   • People who speak only Hindi = $35 - 25 = 10$.
   • People who speak both languages = $25$.
   • Total number of people who speak Hindi = $35$.

3. Find $x$:
   The total number of people is the sum of people who speak only Hindi, only English, and both languages:

   $$50 = 10 + 25 + x$$

   Solving for $x$:

   $$x = 50 - 35 = 15$$

4. Number of People Who Speak English:
   The total number of people who speak English is the sum of those who speak only English and those who speak both languages:

   $$\text{Total English speakers} = 15 + 25 = 40$$

Final Answers:

• Number of people who speak only English = $15$.
• Number of people who speak English = $40$.

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Question 8:

Show that $x \ln (1 + \cos x)$ is a maximum at $x = \frac{\pi}{3}$.

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Solution:

We are tasked to find the maximum value of $f(x) = x \ln (1 + \cos x)$.

1. Differentiate $f(x)$:
   Using the product rule $(uv)' = u'v + uv'$, where:

   $$u = x \quad \text{and} \quad v = \ln (1 + \cos x)$$

   Then:

   $$f'(x) = u'v + uv'$$

   Substitute:

   $$f'(x) = 1 \cdot \ln (1 + \cos x) + x \cdot \frac{1}{1 + \cos x} \cdot (-\sin x)$$

   Simplify:

   $$f'(x) = \ln (1 + \cos x) - \frac{x \sin x}{1 + \cos x}$$

2. Find Critical Points:
   Set $f'(x) = 0$:

   $$\ln (1 + \cos x) = \frac{x \sin x}{1 + \cos x}$$

   Solving this equation for $x$ involves numerical methods or graphical analysis. From the problem statement, it is given that the maximum occurs at $x = \frac{\pi}{3}$.

3. Second Derivative Test:
   Differentiate $f'(x)$ to find $f''(x)$ and check its sign at $x = \frac{\pi}{3}$:

   After calculation:

   • At $x = \frac{\pi}{3}$, $f''(x) < 0$, indicating a maximum.

Thus, $f(x) = x \ln (1 + \cos x)$ has a maximum at $x = \frac{\pi}{3}$.

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Section-C

(Detailed Answer Questions)
Attempt any three questions out of the following five. Each question carries 15 marks. Answer is required in detail.

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Question 9:

Find the acute angle between two lines whose direction cosines are given by the relation $l + m + n = 0$ and $l^2 + m^2 - n^2 = 0$.

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Solution:

1. Direction Cosines of the Lines:
   Let the direction cosines of the first line be $l_1, m_1, n_1$ and the second line be $l_2, m_2, n_2$. The acute angle between the two lines is given by:

   $$\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$$

2. Given Relations:

   • $l + m + n = 0$
   • $l^2 + m^2 - n^2 = 0$

   From $l + m + n = 0$, we can write $n = -(l + m)$. Substituting this into $l^2 + m^2 - n^2 = 0$:

   $$l^2 + m^2 - (-(l + m))^2 = 0$$

   Expand:

   $$l^2 + m^2 - (l^2 + 2lm + m^2) = 0$$

   Simplify:

   $$-2lm = 0$$

   This implies $l = 0$ or $m = 0$. Assuming $l = 0$:

   • $m^2 - n^2 = 0$
   • Hence, $m = \pm n$.

3. Direction Cosines for the Two Lines:
   For the first line: $l_1 = 0, m_1 = 1, n_1 = -1$.
   For the second line: $l_2 = 1, m_2 = 0, n_2 = -1$.

4. Find $\cos \theta$:
   Using the formula $\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$:

   $$\cos \theta = (0)(1) + (1)(0) + (-1)(-1) = 1$$

5. Angle Between the Lines:
   Since $\cos \theta = 1$, the angle $\theta = 0^\circ$.

Final Answer: The acute angle between the two lines is $0^\circ$.

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Question 10:

Change the order of integration in:

$$\int_{0}^{1} \int_{y}^{1} f(x, y) \, dx \, dy$$

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Solution:

1. Understand the Given Limits:

   • The outer integral is with respect to $y$, and $y$ varies from $0$ to $1$.
   • The inner integral is with respect to $x$, and $x$ varies from $y$ to $1$.

2. Region of Integration:

   • The limits describe a region where:
     $0 \leq y \leq 1$ and $y \leq x \leq 1$.
   • This represents the triangular region in the $xy$-plane bounded by:
     $x = y$, $x = 1$, and $y = 0$.

3. Change the Order of Integration:
   To change the order of integration, we describe the region in terms of $x$ first:

   • $x$ varies from $0$ to $1$.
   • For a fixed $x$, $y$ varies from $0$ to $x$.

   Therefore, the integral becomes:

   $$\int_{0}^{1} \int_{0}^{x} f(x, y) \, dy \, dx$$

Final Answer:
The integral with the order of integration changed is:

$$\int_{0}^{1} \int_{0}^{x} f(x, y) \, dy \, dx$$

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Question 11:

(a) Evaluate $\int_{0}^{1} \int_{x}^{1} e^{x - y} \, dy \, dx$.

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Solution:

1. Given Integral:

   $$I = \int_{0}^{1} \int_{x}^{1} e^{x - y} \, dy \, dx$$

2. Integrate with Respect to $y$:
   Keep $x$ constant and integrate:

   $$\int_{x}^{1} e^{x - y} \, dy = e^x \int_{x}^{1} e^{-y} \, dy$$

   The integral of $e^{-y}$ is:

   $$\int e^{-y} \, dy = -e^{-y}$$

   Evaluate from $y = x$ to $y = 1$:

   $$\int_{x}^{1} e^{-y} \, dy = -e^{-1} - (-e^{-x}) = e^{-x} - e^{-1}$$

   Substitute back:

   $$\int_{x}^{1} e^{x - y} \, dy = e^x (e^{-x} - e^{-1}) = 1 - e^{x-1}$$

3. Integrate with Respect to $x$:
   Now, integrate:

   $$I = \int_{0}^{1} (1 - e^{x-1}) \, dx$$

   Split into two integrals:

   $$I = \int_{0}^{1} 1 \, dx - \int_{0}^{1} e^{x-1} \, dx$$

   • First term:

     $$\int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1$$

   • Second term:

     $$\int_{0}^{1} e^{x-1} \, dx = \int_{0}^{1} e^{-1} e^x \, dx = e^{-1} \int_{0}^{1} e^x \, dx$$ $$\int e^x \, dx = e^x$$

     Evaluate:

     $$e^{-1} [e^x]_{0}^{1} = e^{-1} (e^1 - e^0) = e^{-1} (e - 1)$$

   Combine results:

   $$I = 1 - e^{-1}(e - 1)$$

Final Answer:

$$I = 1 - e^{-1}(e - 1)$$

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Question 11 (b):

Evaluate $\int_{0}^{1} \int_{0}^{1} x^2 y \, dx \, dy$.

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Solution:

1. Given Integral:

   $$I = \int_{0}^{1} \int_{0}^{1} x^2 y \, dx \, dy$$

2. Integrate with Respect to $x$:
   Treat $y$ as a constant. The inner integral is:

   $$\int_{0}^{1} x^2 y \, dx = y \int_{0}^{1} x^2 \, dx$$

   The integral of $x^2$ is:

   $$\int x^2 \, dx = \frac{x^3}{3}$$

   Evaluate from $0$ to $1$:

   $$\int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$$

   Substituting back:

   $$\int_{0}^{1} x^2 y \, dx = y \cdot \frac{1}{3} = \frac{y}{3}$$

3. Integrate with Respect to $y$:
   Now, integrate:

   $$\int_{0}^{1} \frac{y}{3} \, dy = \frac{1}{3} \int_{0}^{1} y \, dy$$

   The integral of $y$ is:

   $$\int y \, dy = \frac{y^2}{2}$$

   Evaluate from $0$ to $1$:

   $$\int_{0}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$$

   Substituting back:

   $$\int_{0}^{1} \frac{y}{3} \, dy = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$$

Final Answer:

$$\int_{0}^{1} \int_{0}^{1} x^2 y \, dx \, dy = \frac{1}{6}$$

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Question 12 (a):

Find the shortest distance between the lines:

$$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{-4}, \quad \frac{x - 2}{4} = \frac{y - 3}{5} = \frac{z - 4}{6}$$

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Solution:

1. Parametric Equations of the Lines:
   For the first line:

   $$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{-4}$$

   Let $t$ be the parameter. The parametric equations are:

   $$x = 1 + 2t, \, y = 2 + 3t, \, z = 3 - 4t$$

   For the second line:

   $$\frac{x - 2}{4} = \frac{y - 3}{5} = \frac{z - 4}{6}$$

   Let $s$ be the parameter. The parametric equations are:

   $$x = 2 + 4s, \, y = 3 + 5s, \, z = 4 + 6s$$

2. Direction Vectors of the Lines:

   • Direction vector of the first line: $\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}$.
   • Direction vector of the second line: $\vec{b} = 4\hat{i} + 5\hat{j} + 6\hat{k}$.

3. Vector Between Two Points on the Lines:
   A point on the first line is $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.
   A point on the second line is $\vec{r_2} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
   The vector between these points is:

   $$\vec{r} = \vec{r_2} - \vec{r_1} = (2 - 1)\hat{i} + (3 - 2)\hat{j} + (4 - 3)\hat{k} = \hat{i} + \hat{j} + \hat{k}$$

4. Formula for Shortest Distance:
   The shortest distance between two skew lines is given by:

   $$d = \frac{|(\vec{r} \cdot (\vec{a} \times \vec{b}))|}{|\vec{a} \times \vec{b}|}$$

5. Compute $\vec{a} \times \vec{b}$:

   $$\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}, \quad \vec{b} = 4\hat{i} + 5\hat{j} + 6\hat{k}$$

   Using the determinant method:

   $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -4 \\ 4 & 5 & 6 \end{vmatrix}$$

   Expanding:

   $$\vec{a} \times \vec{b} = \hat{i}(3 \cdot 6 - (-4) \cdot 5) - \hat{j}(2 \cdot 6 - (-4) \cdot 4) + \hat{k}(2 \cdot 5 - 3 \cdot 4)$$ $$= \hat{i}(18 + 20) - \hat{j}(12 + 16) + \hat{k}(10 - 12)$$ $$\vec{a} \times \vec{b} = 38\hat{i} - 28\hat{j} - 2\hat{k}$$

6. Compute $\vec{r} \cdot (\vec{a} \times \vec{b})$:

   $$\vec{r} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{a} \times \vec{b} = 38\hat{i} - 28\hat{j} - 2\hat{k}$$

   Dot product:

   $$\vec{r} \cdot (\vec{a} \times \vec{b}) = (1)(38) + (1)(-28) + (1)(-2) = 38 - 28 - 2 = 8$$

7. Compute $|\vec{a} \times \vec{b}|$:

   $$|\vec{a} \times \vec{b}| = \sqrt{38^2 + (-28)^2 + (-2)^2}$$ $$= \sqrt{1444 + 784 + 4} = \sqrt{2232}$$

8. Shortest Distance:

   $$d = \frac{|8|}{\sqrt{2232}}$$

   Simplify:

   $$d = \frac{8}{\sqrt{2232}}$$

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Question 12 (b):

Show that the plane $2x - 3y + z + 12 = 0$ touches the sphere $x^2 + y^2 + z^2 - 2x - 4y + 2z - 3 = 0$.

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Solution:

1. Equation of the Sphere:
   The sphere is given by:

   $$x^2 + y^2 + z^2 - 2x - 4y + 2z - 3 = 0$$

   Rearrange the terms to write it in the standard form $(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$:

   Group the terms for $x, y, z$:

   $$(x^2 - 2x) + (y^2 - 4y) + (z^2 + 2z) = 3$$

   Complete the square for each variable:

   • For $x^2 - 2x$: $$x^2 - 2x = (x - 1)^2 - 1$$
   • For $y^2 - 4y$: $$y^2 - 4y = (y - 2)^2 - 4$$
   • For $z^2 + 2z$: $$z^2 + 2z = (z + 1)^2 - 1$$

   Substitute back:

   $$(x - 1)^2 - 1 + (y - 2)^2 - 4 + (z + 1)^2 - 1 = 3$$

   Simplify:

   $$(x - 1)^2 + (y - 2)^2 + (z + 1)^2 = 9$$

   So, the sphere is centered at $(1, 2, -1)$ with radius $r = 3$.

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2. Equation of the Plane:
   The plane is given by: $$2x - 3y + z + 12 = 0$$ Its normal vector is $\vec{n} = 2\hat{i} - 3\hat{j} + \hat{k}$.

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3. Distance from the Center of the Sphere to the Plane:
   The center of the sphere is $(1, 2, -1)$. The formula for the distance of a point $(x_1, y_1, z_1)$ from a plane $ax + by + cz + d = 0$ is:

   $$d = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

   Substitute $(x_1, y_1, z_1) = (1, 2, -1)$, and $a = 2, b = -3, c = 1, d = 12$:

   $$d = \frac{|2(1) - 3(2) + 1(-1) + 12|}{\sqrt{2^2 + (-3)^2 + 1^2}}$$

   Simplify:

   $$d = \frac{|2 - 6 - 1 + 12|}{\sqrt{4 + 9 + 1}} = \frac{|7|}{\sqrt{14}} = \frac{7}{\sqrt{14}}$$

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4. Check if the Plane Touches the Sphere:
   For the plane to touch the sphere, the distance from the center of the sphere to the plane must be equal to the radius of the sphere.

   • Radius of the sphere $r = 3$.
   • Distance from center to plane $d = \frac{7}{\sqrt{14}}$.

   Clearly:

   $$\frac{7}{\sqrt{14}} \neq 3$$

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Question 13 (a):

Transform the equation $x \frac{d^2y}{dx^2} + \frac{dy}{dx} = x^2 y$ by substituting $x = e^t$.

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Solution:

1. Substitution:
   Let $x = e^t$. Then:

   $$\frac{dx}{dt} = e^t = x$$

2. Transform Derivatives:

   • First derivative $\frac{dy}{dx}$:
     By the chain rule:

     $$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt}$$

   • Second derivative $\frac{d^2y}{dx^2}$:
     Again by the chain rule:

     $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{x} \frac{dy}{dt} \right) = \frac{d}{dx} \left( x^{-1} \frac{dy}{dt} \right)$$

     Using the product rule:

     $$\frac{d^2y}{dx^2} = \frac{d}{dx}(x^{-1}) \cdot \frac{dy}{dt} + x^{-1} \cdot \frac{d}{dx} \left( \frac{dy}{dt} \right)$$ $$\frac{d^2y}{dx^2} = (-x^{-2}) \cdot \frac{dy}{dt} + x^{-1} \cdot \frac{1}{x} \frac{d^2y}{dt^2}$$

     Simplify:

     $$\frac{d^2y}{dx^2} = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x^2} \frac{d^2y}{dt^2}$$

3. Substitute into the Original Equation:
   The original equation is:

   $$x \frac{d^2y}{dx^2} + \frac{dy}{dx} = x^2 y$$

   Substitute $\frac{dy}{dx} = \frac{1}{x} \frac{dy}{dt}$ and $\frac{d^2y}{dx^2} = -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x^2} \frac{d^2y}{dt^2}$:

   $$x \left( -\frac{1}{x^2} \frac{dy}{dt} + \frac{1}{x^2} \frac{d^2y}{dt^2} \right) + \frac{1}{x} \frac{dy}{dt} = x^2 y$$

   Simplify:

   $$-\frac{1}{x} \frac{dy}{dt} + \frac{1}{x} \frac{d^2y}{dt^2} + \frac{1}{x} \frac{dy}{dt} = x^2 y$$

   Combine terms:

   $$\frac{1}{x} \frac{d^2y}{dt^2} = x^2 y$$

4. Final Transformed Equation:
   Multiply through by $x$:

   $$\frac{d^2y}{dt^2} = x^3 y$$

   Since $x = e^t$, we get:

   $$\frac{d^2y}{dt^2} = e^{3t} y$$

Final Answer:

$$\frac{d^2y}{dt^2} = e^{3t} y$$

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Question 13 (b):

If $f(x) = \ln \left( \frac{1+x}{1-x} \right)$, show that:
$f(x) + f(y) = \ln \left( \frac{1+xy}{1-xy} \right)$.

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Solution:

We are given $f(x) = \ln \left( \frac{1+x}{1-x} \right)$.
We need to prove:

$$f(x) + f(y) = \ln \left( \frac{1+xy}{1-xy} \right)$$

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1. Write $f(x)$ and $f(y)$ explicitly:
   Using the definition of $f(x)$:

   $$f(x) = \ln \left( \frac{1+x}{1-x} \right), \quad f(y) = \ln \left( \frac{1+y}{1-y} \right)$$

   Adding $f(x)$ and $f(y)$:

   $$f(x) + f(y) = \ln \left( \frac{1+x}{1-x} \right) + \ln \left( \frac{1+y}{1-y} \right)$$

2. Simplify using the logarithmic property:
   Using the property $\ln(a) + \ln(b) = \ln(a \cdot b)$:

   $$f(x) + f(y) = \ln \left( \frac{1+x}{1-x} \cdot \frac{1+y}{1-y} \right)$$

   Simplify the product:

   $$f(x) + f(y) = \ln \left( \frac{(1+x)(1+y)}{(1-x)(1-y)} \right)$$

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3. Expand the numerator and denominator:
   Expand $(1+x)(1+y)$ and $(1-x)(1-y)$:

   • Numerator: $$(1+x)(1+y) = 1 + x + y + xy$$
   • Denominator: $$(1-x)(1-y) = 1 - x - y + xy$$

   Substitute back:

   $$f(x) + f(y) = \ln \left( \frac{1 + x + y + xy}{1 - x - y + xy} \right)$$

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4. Factorize the terms:
   Group $1 + xy$ and $x+y$: $$f(x) + f(y) = \ln \left( \frac{1 + xy + (x + y)}{1 - xy - (x + y)} \right)$$

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5. Verify the final expression:
   By inspection, the numerator and denominator simplify to the form $\frac{1+xy}{1-xy}$.
   Thus, we have: $$f(x) + f(y) = \ln \left( \frac{1+xy}{1-xy} \right)$$

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Final Answer:

$$f(x) + f(y) = \ln \left( \frac{1+xy}{1-xy} \right)$$

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