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Unit2 - Subjective Questions

MTH174 • Practice Questions with Detailed Answers

Define a Linear Differential Equation (LDE). What are its general characteristics?

A Linear Differential Equation (LDE) is a differential equation in which the dependent variable and its derivatives appear only in the first degree, and they are not multiplied together.

The general form of an $n$ -th order linear differential equation is:
$a_n(x)\\frac{d^ny}{dx^n} + a_{n-1}(x)\\frac{d^{n-1}y}{dx^{n-1}} + \\dots + a_1(x)\\frac{dy}{dx} + a_0(x)y = f(x)$

Characteristics:

Degree: The degree of the dependent variable $y$ and all its derivatives is exactly 1.
No Products: There are no terms involving the product of $y$ and its derivatives (e.g., $y \\frac{dy}{dx}$ is not allowed).
No Non-linear Functions: The dependent variable $y$ does not appear inside trigonometric, exponential, or logarithmic functions (e.g., $\\sin(y)$ or $e^y$ are not allowed).
If $f(x) = 0$ , the equation is called homogeneous. If $f(x) \\neq 0$ , it is called non-homogeneous.

Distinguish between homogeneous and non-homogeneous linear differential equations.

Homogeneous Linear Differential Equations:

An LDE is homogeneous if the right-hand side function $f(x)$ is identically zero.
General form: $a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \\dots + a_0(x)y = 0$ .
The solution consists only of the Complementary Function (CF).
It obeys the principle of superposition: if $y_1$ and $y_2$ are solutions, then $c_1y_1 + c_2y_2$ is also a solution.

Non-homogeneous Linear Differential Equations:

An LDE is non-homogeneous if the right-hand side function $f(x)$ is not zero.
General form: $a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \\dots + a_0(x)y = f(x)$ .
The complete solution is the sum of the Complementary Function (CF) and the Particular Integral (PI).
It does not obey the simple principle of superposition for its complete solution.

Explain the concept of linear dependence and linear independence of solutions using the Wronskian.

Let $y_1(x), y_2(x), \\dots, y_n(x)$ be $n$ solutions to an $n$ -th order linear differential equation.

Linear Dependence:
The functions are linearly dependent if there exist constants $c_1, c_2, \\dots, c_n$ (not all zero) such that:
$c_1y_1(x) + c_2y_2(x) + \\dots + c_ny_n(x) = 0$
for all $x$ in the given interval.

Linear Independence:
The functions are linearly independent if the above equation holds true only when $c_1 = c_2 = \\dots = c_n = 0$ .

The Wronskian ( $W$ ):
Linear independence can be tested using the Wronskian determinant:
$W(y_1, y_2, \\dots, y_n) = \\begin{vmatrix} y_1 & y_2 & \\dots & y_n \\ y_1' & y_2' & \\dots & y_n' \\ \\vdots & \\vdots & \\ddots & \\vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \\dots & y_n^{(n-1)} \\end{vmatrix}$

If $W \\neq 0$ for at least one point in the interval, the solutions are linearly independent.
If $W = 0$ everywhere in the interval, the solutions are linearly dependent.

Check the linear independence of the functions $y_1 = e^x$ and $y_2 = e^{-x}$ .

To check the linear independence of $y_1 = e^x$ and $y_2 = e^{-x}$ , we calculate their Wronskian $W(y_1, y_2)$ .

Step 1: Find the derivatives.
$y_1 = e^x \\implies y_1' = e^x$
$y_2 = e^{-x} \\implies y_2' = -e^{-x}$

Step 2: Construct the Wronskian determinant.
$W(y_1, y_2) = \\begin{vmatrix} e^x & e^{-x} \\ e^x & -e^{-x} \\end{vmatrix}$

Step 3: Evaluate the determinant.
$W = (e^x)(-e^{-x}) - (e^{-x})(e^x)$
$W = -e^0 - e^0$
$W = -1 - 1 = -2$

Since $W = -2 \\neq 0$ , the functions $y_1 = e^x$ and $y_2 = e^{-x}$ are linearly independent.

Define the Differential Operator $D$ and list its basic properties.

The Differential Operator, denoted by $D$ , is a shorthand notation used to represent the derivative with respect to an independent variable (usually $x$ ).

Definition:
$D \\equiv \\frac{d}{dx}$
Hence, $Dy = \\frac{dy}{dx}$ , $D^2y = \\frac{d^2y}{dx^2}$ , and $D^ny = \\frac{d^ny}{dx^n}$ .

A linear differential equation with constant coefficients can be written as:
$(a_n D^n + a_{n-1} D^{n-1} + \\dots + a_1 D + a_0)y = f(x)$
or $f(D)y = f(x)$ .

Properties of Operator $D$ :

Linearity: $D[af(x) + bg(x)] = aDf(x) + bDg(x)$ , where $a, b$ are constants.
Laws of Algebra: The operator polynomial with constant coefficients satisfies the commutative, associative, and distributive laws of ordinary algebra.
- $D^m(D^n y) = D^{m+n} y$
- $(D - a)(D - b)y = (D - b)(D - a)y$
Inverse Operator: $\\frac{1}{D}$ represents integration with respect to $x$ . $\\frac{1}{D} f(x) = \\int f(x) dx$ .

Explain the method of finding the general solution for a second-order homogeneous linear differential equation with constant coefficients.

A second-order homogeneous linear differential equation with constant coefficients has the form:
$a\\frac{d^2y}{dx^2} + b\\frac{dy}{dx} + cy = 0$
or $(aD^2 + bD + c)y = 0$ , where $a, b, c$ are constants.

Method of Solution:

Assume a trial solution: Let $y = e^{mx}$ . Then $Dy = me^{mx}$ and $D^2y = m^2e^{mx}$ .
Substitute into the equation:
$a(m^2e^{mx}) + b(me^{mx}) + c(e^{mx}) = 0$
$e^{mx}(am^2 + bm + c) = 0$
Form the Auxiliary Equation (AE): Since $e^{mx} \\neq 0$ , we have:
$am^2 + bm + c = 0$
Find the roots of the AE: Let the roots be $m_1$ and $m_2$ . The general solution depends on the nature of these roots:
- Case 1: Roots are real and distinct ( $m_1 \\neq m_2$ ).
  General Solution: $y = c_1e^{m_1x} + c_2e^{m_2x}$
- Case 2: Roots are real and equal ( $m_1 = m_2 = m$ ).
  General Solution: $y = (c_1 + c_2x)e^{mx}$
- Case 3: Roots are complex conjugates ( $m = \\alpha \\pm i\\beta$ ).
  General Solution: $y = e^{\\alpha x}(c_1\\cos\\beta x + c_2\\sin\\beta x)$

Solve the differential equation: $\\frac{d^2y}{dx^2} - 5\\frac{dy}{dx} + 6y = 0$ .

Step 1: Write the equation using the D operator.
$(D^2 - 5D + 6)y = 0$

Step 2: Write the Auxiliary Equation (AE).
Replace $D$ with $m$ :
$m^2 - 5m + 6 = 0$

Step 3: Solve the AE for $m$ .
Factorize the quadratic equation:
$m^2 - 3m - 2m + 6 = 0$
$m(m - 3) - 2(m - 3) = 0$
$(m - 2)(m - 3) = 0$

The roots are $m_1 = 2$ and $m_2 = 3$ .

Step 4: Write the general solution.
Since the roots are real and distinct, the general solution is:
$y = c_1e^{m_1x} + c_2e^{m_2x}$
$y = c_1e^{2x} + c_2e^{3x}$
where $c_1$ and $c_2$ are arbitrary constants.

Find the general solution of the differential equation $y'' - 4y' + 4y = 0$ .

Step 1: Write the equation in Operator form.
$(D^2 - 4D + 4)y = 0$

Step 2: Form the Auxiliary Equation.
$m^2 - 4m + 4 = 0$

Step 3: Find the roots.
This is a perfect square:
$(m - 2)^2 = 0$
$m = 2, 2$

The roots are real and equal ( $m_1 = m_2 = 2$ ).

Step 4: Write the general solution.
For real and repeated roots, the general solution is of the form:
$y = (c_1 + c_2x)e^{mx}$
Substituting $m = 2$ :
$y = (c_1 + c_2x)e^{2x}$
where $c_1$ and $c_2$ are arbitrary constants.

Find the general solution to $y'' + 4y = 0$ .

Step 1: Write the equation in Operator form.
$(D^2 + 4)y = 0$

Step 2: Form the Auxiliary Equation.
$m^2 + 4 = 0$

Step 3: Solve for $m$ .
$m^2 = -4$
$m = \\pm \\sqrt{-4}$
$m = 0 \\pm 2i$

The roots are complex conjugates of the form $\\alpha \\pm i\\beta$ , where $\\alpha = 0$ and $\\beta = 2$ .

Step 4: Write the general solution.
For complex roots $\\alpha \\pm i\\beta$ , the solution is $y = e^{\\alpha x}(c_1\\cos\\beta x + c_2\\sin\\beta x)$ .
Substituting $\\alpha = 0$ and $\\beta = 2$ :
$y = e^{0 \\cdot x}(c_1\\cos 2x + c_2\\sin 2x)$
$y = c_1\\cos 2x + c_2\\sin 2x$
where $c_1$ and $c_2$ are arbitrary constants.

Explain the general solution structure for an $n$ -th order homogeneous linear differential equation with constant coefficients.

An $n$ -th order homogeneous LDE with constant coefficients is given by:
$(a_n D^n + a_{n-1} D^{n-1} + \\dots + a_1 D + a_0)y = 0$

The Auxiliary Equation (AE) is a polynomial of degree $n$ :
$a_n m^n + a_{n-1} m^{n-1} + \\dots + a_1 m + a_0 = 0$

By the Fundamental Theorem of Algebra, this AE has exactly $n$ roots (real or complex, distinct or repeated). The general solution $y$ is the sum of the solutions corresponding to each root.

Rules for forming the solution:

Real and Distinct Roots: If $m_1, m_2, \\dots, m_k$ are distinct real roots, the corresponding part of the solution is:
$y = c_1e^{m_1x} + c_2e^{m_2x} + \\dots + c_ke^{m_kx}$
Real and Repeated Roots: If a real root $m$ is repeated $r$ times, the corresponding part of the solution is:
$y = (c_1 + c_2x + c_3x^2 + \\dots + c_rx^{r-1})e^{mx}$
Complex Distinct Roots: If there is a pair of complex roots $\\alpha \\pm i\\beta$ , the corresponding part of the solution is:
$y = e^{\\alpha x}(c_1\\cos\\beta x + c_2\\sin\\beta x)$
Complex Repeated Roots: If a complex pair $\\alpha \\pm i\\beta$ is repeated $r$ times, the corresponding part of the solution is:
$y = e^{\\alpha x}((c_1 + c_2x + \\dots + c_rx^{r-1})\\cos\\beta x + (d_1 + d_2x + \\dots + d_rx^{r-1})\\sin\\beta x)$

The total general solution contains exactly $n$ arbitrary constants.

Solve the higher order differential equation: $\\frac{d^3y}{dx^3} - 6\\frac{d^2y}{dx^2} + 11\\frac{dy}{dx} - 6y = 0$ .

Step 1: Write in Operator form.
$(D^3 - 6D^2 + 11D - 6)y = 0$

Step 2: Form the Auxiliary Equation.
$m^3 - 6m^2 + 11m - 6 = 0$

Step 3: Find the roots.
By trial and error, test factors of 6. Let $m = 1$ :
$(1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$ .
So, $m=1$ is a root, and $(m-1)$ is a factor.
Using synthetic division or polynomial division to divide $m^3 - 6m^2 + 11m - 6$ by $(m-1)$ :
Quotient is $m^2 - 5m + 6$ .
So, the AE becomes $(m - 1)(m^2 - 5m + 6) = 0$ .
Factorizing the quadratic part:
$m^2 - 5m + 6 = (m - 2)(m - 3)$ .
Thus, $(m - 1)(m - 2)(m - 3) = 0$ .

The roots are $m_1 = 1$ , $m_2 = 2$ , $m_3 = 3$ .

Step 4: Write the general solution.
Since the roots are real and distinct:
$y = c_1e^{x} + c_2e^{2x} + c_3e^{3x}$
where $c_1, c_2, c_3$ are arbitrary constants.

Prove the principle of superposition for homogeneous linear differential equations.

Let $L(y) = 0$ represent an $n$ -th order linear homogeneous differential equation:
$L(y) = a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \\dots + a_0(x)y = 0$

Proof of Superposition Principle:
Assume $y_1(x)$ and $y_2(x)$ are two solutions of $L(y) = 0$ .
This means $L(y_1) = 0$ and $L(y_2) = 0$ .

We need to prove that $y = c_1y_1 + c_2y_2$ is also a solution, where $c_1$ and $c_2$ are arbitrary constants.

Apply the linear operator $L$ to $y$ :
$L(c_1y_1 + c_2y_2) = a_n(x)(c_1y_1 + c_2y_2)^{(n)} + \\dots + a_0(x)(c_1y_1 + c_2y_2)$

Because the derivative operator is linear, $(c_1y_1 + c_2y_2)^{(k)} = c_1y_1^{(k)} + c_2y_2^{(k)}$ .
Rearranging the terms by factoring out $c_1$ and $c_2$ :
$L(c_1y_1 + c_2y_2) = c_1 [a_n(x)y_1^{(n)} + \\dots + a_0(x)y_1] + c_2 [a_n(x)y_2^{(n)} + \\dots + a_0(x)y_2]$
$L(c_1y_1 + c_2y_2) = c_1 L(y_1) + c_2 L(y_2)$

Since $L(y_1) = 0$ and $L(y_2) = 0$ , we have:
$L(c_1y_1 + c_2y_2) = c_1(0) + c_2(0) = 0$

Therefore, $c_1y_1 + c_2y_2$ is also a solution. This confirms the principle of superposition for homogeneous LDEs.

What is meant by the Complementary Function (CF) in the context of linear differential equations?

The Complementary Function (CF) is the general solution to the corresponding homogeneous part of a linear differential equation.

Given a non-homogeneous linear differential equation:
$f(D)y = X$
where $X$ is a function of $x$ .

The corresponding homogeneous equation is obtained by setting $X = 0$ :
$f(D)y = 0$

The general solution of $f(D)y = 0$ is called the Complementary Function.
It contains a number of arbitrary constants equal to the order of the differential equation.
It completely describes the behavior of the system without the external forcing function $X$ .
For a homogeneous LDE ( $X=0$ ), the complete solution is purely the CF ( $y = \\text{CF}$ ). For a non-homogeneous LDE, the general solution is the sum of the CF and the Particular Integral (PI).

Solve the differential equation: $(D^3 - 3D^2 + 3D - 1)y = 0$ .

Step 1: Write the Auxiliary Equation (AE).
$m^3 - 3m^2 + 3m - 1 = 0$

Step 2: Solve the AE for $m$ .
We recognize the left-hand side as the algebraic expansion of a perfect cube:
$(m - 1)^3 = m^3 - 3m^2(1) + 3m(1^2) - 1^3$
So, the AE is:
$(m - 1)^3 = 0$

The roots are $m = 1, 1, 1$ . This is a real root repeated three times ( $r=3$ ).

Step 3: Write the general solution.
For a real root $m$ repeated 3 times, the solution is of the form:
$y = (c_1 + c_2x + c_3x^2)e^{mx}$

Substituting $m=1$ :
$y = (c_1 + c_2x + c_3x^2)e^x$
where $c_1, c_2, c_3$ are arbitrary constants.

State the necessary and sufficient condition for $n$ solutions to be linearly independent over an interval.

Let $y_1, y_2, \\dots, y_n$ be $n$ solutions to an $n$ -th order homogeneous linear differential equation $L(y) = 0$ on an interval $I$ .

Condition:
The necessary and sufficient condition for these $n$ solutions to be linearly independent on the interval $I$ is that their Wronskian determinant $W(y_1, y_2, \\dots, y_n)$ is non-zero at least at one point in the interval $I$ .

Mathematically:
$W(y_1, y_2, \\dots, y_n) \\neq 0$

Because of Abel's Identity, if the Wronskian of the solutions to a linear ODE is non-zero at one point in the interval, it is non-zero at all points in the interval. If it is zero at one point, it is zero everywhere. Thus, checking $W \\neq 0$ at any convenient point $x_0 \\in I$ is sufficient to prove linear independence.

Solve the fourth-order differential equation: $(D^4 - 16)y = 0$ .

Step 1: Write the Auxiliary Equation.
$m^4 - 16 = 0$

Step 2: Find the roots.
Factorize as a difference of squares:
$(m^2 - 4)(m^2 + 4) = 0$

This gives two equations:
1) $m^2 - 4 = 0 \\implies m^2 = 4 \\implies m = \\pm 2$
2) $m^2 + 4 = 0 \\implies m^2 = -4 \\implies m = \\pm 2i$

The roots are $m_1 = 2$ , $m_2 = -2$ (real and distinct), and $m_{3,4} = 0 \\pm 2i$ (complex conjugates with $\\alpha = 0, \\beta = 2$ ).

Step 3: Write the general solution.
Combine the solutions for the real parts and complex parts:
$y = c_1e^{2x} + c_2e^{-2x} + e^{0x}(c_3\\cos 2x + c_4\\sin 2x)$

Simplifying:
$y = c_1e^{2x} + c_2e^{-2x} + c_3\\cos 2x + c_4\\sin 2x$
where $c_1, c_2, c_3, c_4$ are arbitrary constants.

Describe how to determine the general solution when the auxiliary equation yields complex repeated roots.

When the auxiliary equation of an $n$ -th order linear homogeneous differential equation with constant coefficients yields complex roots, they always occur in conjugate pairs (assuming real coefficients).

If a complex conjugate pair $m = \\alpha \\pm i\\beta$ is repeated $r$ times, the corresponding roots are $\\alpha \\pm i\\beta, \\alpha \\pm i\\beta, \\dots$ ( $r$ times).

General Solution Structure:
Just as repeated real roots require multiplying by powers of $x$ to ensure linear independence ( $e^{mx}, xe^{mx}, x^2e^{mx}$ ), repeated complex roots require the same treatment.

The independent solutions are:
$e^{\\alpha x}\\cos\\beta x, \\quad xe^{\\alpha x}\\cos\\beta x, \\quad \\dots, \\quad x^{r-1}e^{\\alpha x}\\cos\\beta x$
$e^{\\alpha x}\\sin\\beta x, \\quad xe^{\\alpha x}\\sin\\beta x, \\quad \\dots, \\quad x^{r-1}e^{\\alpha x}\\sin\\beta x$

Combining these using the principle of superposition, the general solution part corresponding to these roots is:
$y = e^{\\alpha x} \\left[ (c_1 + c_2x + \\dots + c_rx^{r-1})\\cos\\beta x + (d_1 + d_2x + \\dots + d_rx^{r-1})\\sin\\beta x \\right]$
where $c_i$ and $d_i$ are $2r$ arbitrary constants.

Solve the second-order differential equation $y'' + 2y' + 5y = 0$ .

Step 1: Write the equation in Operator form.
$(D^2 + 2D + 5)y = 0$

Step 2: Form the Auxiliary Equation.
$m^2 + 2m + 5 = 0$

Step 3: Find the roots.
Using the quadratic formula $m = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$ :
$m = \\frac{-2 \\pm \\sqrt{4 - 4(1)(5)}}{2}$
$m = \\frac{-2 \\pm \\sqrt{4 - 20}}{2}$
$m = \\frac{-2 \\pm \\sqrt{-16}}{2}$
$m = \\frac{-2 \\pm 4i}{2}$
$m = -1 \\pm 2i$

The roots are complex conjugates with $\\alpha = -1$ and $\\beta = 2$ .

Step 4: Write the general solution.
The solution is $y = e^{\\alpha x}(c_1\\cos\\beta x + c_2\\sin\\beta x)$ .
Substituting $\\alpha = -1$ and $\\beta = 2$ :
$y = e^{-x}(c_1\\cos 2x + c_2\\sin 2x)$
where $c_1$ and $c_2$ are arbitrary constants.

Explain the significance of the roots of the auxiliary equation in determining the stability of a physical system modeled by an LDE.

In engineering, homogeneous linear differential equations often model unforced physical systems (like RLC circuits or mass-spring-damper systems). The solution represents the natural response of the system over time $t$ .

The general solution is a sum of exponential terms $e^{mt}$ , where $m$ are the roots of the auxiliary equation. The real part of these roots, $\\alpha$ , dictates the system's stability:

Stable System (Overdamped/Underdamped): If all roots have negative real parts ( $\\alpha < 0$ ), the terms $e^{\\alpha t}$ decay to zero as $t \\to \\infty$ . The system naturally settles down to an equilibrium state.
Unstable System: If any root has a positive real part ( $\\alpha > 0$ ), the term $e^{\\alpha t}$ grows exponentially as $t \\to \\infty$ . The system's response unbounded.
Marginally Stable (Oscillatory): If roots are purely imaginary (complex with zero real part, $\\alpha = 0$ ) and distinct, the solution contains pure sine and cosine terms. The system oscillates continuously with constant amplitude (e.g., an undamped pendulum).

Thus, the roots of the auxiliary equation directly characterize the transient behavior and stability of the system.

Solve the equation $(D^3 - D^2 - D + 1)y = 0$ .

Step 1: Form the Auxiliary Equation.
$m^3 - m^2 - m + 1 = 0$

Step 2: Solve the AE.
Group the terms to factorize:
$m^2(m - 1) - 1(m - 1) = 0$
$(m^2 - 1)(m - 1) = 0$

Expand the difference of squares:
$(m - 1)(m + 1)(m - 1) = 0$
$(m - 1)^2 (m + 1) = 0$

The roots are $m_1 = 1$ , $m_2 = 1$ , and $m_3 = -1$ .
This means $m=1$ is a repeated real root (multiplicity 2), and $m=-1$ is a distinct real root.

Step 3: Write the general solution.
For the repeated root $m=1$ , the solution part is $(c_1 + c_2x)e^x$ .
For the distinct root $m=-1$ , the solution part is $c_3e^{-x}$ .

Combining them:
$y = (c_1 + c_2x)e^x + c_3e^{-x}$
where $c_1, c_2, c_3$ are arbitrary constants.

Unit1 Unit3