(SEM III) THEORY EXAMINATION 2018-19 MATHEMATICS-III

Here are two selected questions typical of a Mathematics-III (Engineering/B.Tech) Theory Examination (circa 2018-19). These usually cover topics like Complex Analysis, Integral Transforms (Laplace/Fourier), and Numerical Techniques.

I have selected one question from Complex Analysis and one from Laplace Transforms, as these are high-weightage sections in almost every Math-III syllabus.

Question 1: Analytic Functions (Complex Analysis)

The Question:
Show that the function $u(x, y) = x^3 - 3xy^2$ is harmonic.

Type of Question:

This is a Complex Analysis / Partial Differential Equation problem. It specifically tests the understanding of Harmonic Functions and the Laplace Equation.

What this is related to:

This concept is fundamental in physics and engineering, particularly in Fluid Dynamics (fluid flow potentials) and Electrostatics (electric potential fields). If a function represents a potential field, it must satisfy the Laplace equation (be harmonic).ShutterstockExplore

Full Explanation & Solution:

Step 1: Understand the Condition

A real-valued function $u(x, y)$ is called Harmonic if it satisfies the Laplace Equation:

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$

We need to find the second-order partial derivatives of the given function with respect to $x$ and $y$ and see if they add up to zero.

Step 2: Calculate Partial Derivatives with respect to $x$

Given: $u = x^3 - 3xy^2$

First derivative ($\frac{\partial u}{\partial x}$): Treat $y$ as a constant.

$$\frac{\partial u}{\partial x} = 3x^2 - 3y^2$$

Second derivative ($\frac{\partial^2 u}{\partial x^2}$): Differentiate the result again with respect to $x$.

$$\frac{\partial^2 u}{\partial x^2} = 6x$$

Step 3: Calculate Partial Derivatives with respect to $y$

First derivative ($\frac{\partial u}{\partial y}$): Treat $x$ as a constant.

$$\frac{\partial u}{\partial y} = 0 - 3x(2y) = -6xy$$

Second derivative ($\frac{\partial^2 u}{\partial y^2}$): Differentiate the result again with respect to $y$.

$$\frac{\partial^2 u}{\partial y^2} = -6x$$

Step 4: Verify the Laplace Equation

Now, add the two second derivatives:

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 6x + (-6x) = 0$$

Conclusion:

Since the sum is zero, the function satisfies the Laplace equation. Therefore, $u(x, y)$ is a harmonic function.

Question 2: Inverse Laplace Transform

The Question
Find the Inverse Laplace Transform of:
$$F(s) = \frac{s+1}{s^2 + 2s + 5}$$

Type of Question:

This is an Integral Transform problem. It tests the ability to manipulate algebraic expressions in the frequency domain ($s$-domain) to return them to the time domain ($t$-domain) using the First Shifting Theorem.

What this is related to:

Laplace transforms are the backbone of Control Systems and Circuit Analysis. This specific math allows engineers to solve complex differential equations (like mass-spring-damper systems or RLC circuits) by turning them into simple algebra problems.

Full Explanation & Solution:

Step 1: Analyze the Denominator

The denominator is $s^2 + 2s + 5$. This cannot be factored into real linear factors easily (roots are complex). Therefore, we must complete the square.

Compare $s^2 + 2s$ with $(s+a)^2 = s^2 + 2as + a^2$.

Here, $2a = 2$, so $a = 1$.

$$s^2 + 2s + 5 = (s^2 + 2s + 1) + 4 = (s+1)^2 + 2^2$$

Step 2: Rewrite the Function

Now substitute this back into $F(s)$:

$$L^{-1} \left\{ \frac{s+1}{(s+1)^2 + 2^2} \right\}$$

Step 3: Apply the First Shifting Theorem

The First Shifting Theorem states that:

$$L^{-1} \{ F(s+a) \} = e^{-at} L^{-1} \{ F(s) \}$$

Here, we have a shift of $+1$ (i.e., $a=1$). We can "pull out" this shift as an exponential term $e^{-1t}$.

Once we remove the shift $(s+1) \to s$, the expression becomes:

$$e^{-t} L^{-1} \left\{ \frac{s}{s^2 + 2^2} \right\}$$

Step 4: Apply Standard Formulas

We recognize the standard Laplace form:

$$L^{-1} \left\{ \frac{s}{s^2 + k^2} \right\} = \cos(kt)$$

In our case, $k = 2$.

Step 5: Final Result

Combining the exponential term and the cosine term:

$$f(t) = e^{-t} \cos(2t)$$

Conclusion:

The inverse Laplace transform is $e^{-t} \cos(2t)$.