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Unit2 - Subjective Questions

MTH166 • Practice Questions with Detailed Answers

Define a Linear Differential Equation of order $n$ . Distinguish it from a non-linear differential equation.

Definition:
A differential equation is said to be linear if:

The dependent variable (say $y$ ) and all its derivatives occur in the first degree only.
No products of the dependent variable and its derivatives (e.g., $y \cdot \frac{dy}{dx}$ ) occur.
No transcendental functions of the dependent variable exist.

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

Distinction:
If any of the above conditions are violated (e.g., presence of terms like $y^2$ , $\sin(y)$ , or $y\frac{dy}{dx}$ ), the equation is non-linear.

Explain the concept of Linear Dependence and Linear Independence of solutions using the Wronskian.

Let $y_1, y_2, \dots, y_n$ be $n$ solutions of a linear differential equation.

Linear Independence:
The functions are linearly independent on an interval $I$ if the equation $c_1 y_1 + c_2 y_2 + \dots + c_n y_n = 0$ implies that all constants $c_1 = c_2 = \dots = c_n = 0$ .

Wronskian ( $W$ ):
The Wronskian of these functions is the determinant:
$W(y_1, \dots, y_n) = \begin{vmatrix} y_1 & y_2 & \dots & y_n \\ y_1' & y_2' & \dots & y_n' \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \dots & y_n^{(n-1)} \end{vmatrix}$

Condition:

If $W \neq 0$ for at least one point in the interval, the functions are Linearly Independent.
If $W = 0$ identically over the interval, the functions are Linearly Dependent.

Verify whether the functions $e^x$ and $e^{2x}$ are linearly independent solutions of a differential equation.

To check for linear independence, we calculate the Wronskian $W$ of $y_1 = e^x$ and $y_2 = e^{2x}$ .

First derivatives:
$y_1' = e^x$
$y_2' = 2e^{2x}$

The Wronskian is given by:
$W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} e^x & e^{2x} \\ e^x & 2e^{2x} \end{vmatrix}$

$W = (e^x)(2e^{2x}) - (e^{2x})(e^x)$
$W = 2e^{3x} - e^{3x}$
$W = e^{3x}$

Since $e^{3x} \neq 0$ for any real $x$ , the Wronskian is non-zero. Therefore, $e^x$ and $e^{2x}$ are linearly independent.

State the Principle of Superposition for homogeneous linear differential equations.

Principle of Superposition:

If $y_1(x)$ and $y_2(x)$ are two solutions of a homogeneous linear differential equation:
$a_n(x)y^{(n)} + \dots + a_0(x)y = 0$

Then, any linear combination of these solutions:
$y = c_1 y_1(x) + c_2 y_2(x)$
where $c_1$ and $c_2$ are arbitrary constants, is also a solution of the differential equation.

Generalization:
If $y_1, y_2, \dots, y_n$ are $n$ linearly independent solutions, then the general solution is $y = c_1 y_1 + c_2 y_2 + \dots + c_n y_n$ .

Explain the role of the Differential Operator ( $D$ ) in solving linear differential equations. What does $\frac{1}{f(D)}X$ represent?

Differential Operator $D$ :
We denote $\frac{d}{dx}$ by $D$ , $\frac{d^2}{dx^2}$ by $D^2$ , and so on. A linear differential equation with constant coefficients can be written as a polynomial in $D$ :
$f(D)y = Q(x)$
where $f(D) = a_n D^n + a_{n-1} D^{n-1} + \dots + a_0$ .

Inverse Operator:
The symbol $\frac{1}{f(D)}$ represents the inverse operation of differentiation, i.e., integration. Specifically, $\frac{1}{f(D)}Q(x)$ represents the Particular Integral (P.I.) of the differential equation, which is a function that, when operated on by $f(D)$ , yields $Q(x)$ .

Outline the general procedure to find the solution of a Homogeneous Linear Differential Equation with Constant Coefficients.

Consider the equation $a_n \frac{d^ny}{dx^n} + \dots + a_0 y = 0$ .

Procedure:

Write the differential equation in operator form: $f(D)y = 0$ .
Form the Auxiliary Equation (A.E.): Replace $D$ with $m$ to get the algebraic equation $f(m) = 0$ .
Find the roots: Solve $f(m) = 0$ for roots $m_1, m_2, \dots, m_n$ .
Write the General Solution based on the nature of roots:
- Real and Distinct ( $m_1, m_2$ ): $y = c_1 e^{m_1 x} + c_2 e^{m_2 x}$
- Real and Equal ( $m_1 = m_2 = m$ ): $y = (c_1 + c_2 x)e^{mx}$
- Complex Conjugate ( $\,\alpha \pm i\beta\,$ ): $y = e^{\alpha x}(c_1 \cos \beta x + c_2 \sin \beta x)$

Solve the following differential equation: $\frac{d^2y}{dx^2} - 7\frac{dy}{dx} + 12y = 0$

Step 1: Write the Auxiliary Equation (A.E.)
Replacing $\frac{d}{dx}$ with $m$ , we get:
$m^2 - 7m + 12 = 0$

Step 2: Find the roots
Factorize the quadratic equation:
$(m - 3)(m - 4) = 0$
Roots are $m_1 = 3$ and $m_2 = 4$ .

Step 3: Write the solution
Since roots are real and distinct, the general solution is:
$y = c_1 e^{3x} + c_2 e^{4x}$

Find the general solution of the differential equation: $(D^2 + 6D + 9)y = 0$

Step 1: Auxiliary Equation
The A.E. is given by:
$m^2 + 6m + 9 = 0$

Step 2: Solve for m
$(m + 3)^2 = 0$
Roots are $m = -3, -3$ (Real and repeated).

Step 3: General Solution
For repeated real roots $m$ , the solution is of the form $y = (c_1 + c_2 x)e^{mx}$ .
Substituting $m = -3$ :
$y = (c_1 + c_2 x)e^{-3x}$

Solve the homogeneous differential equation: $y'' + 4y = 0$

Step 1: Auxiliary Equation
$m^2 + 4 = 0$

Step 2: Solve for roots
$m^2 = -4$
$m = \pm \sqrt{-4}$
$m = 0 \pm 2i$
The roots are complex conjugates with $\alpha = 0$ and $\beta = 2$ .

Step 3: General Solution
The solution form is $y = e^{\alpha x}(c_1 \cos \beta x + c_2 \sin \beta x)$ .
Substituting $\alpha=0, \beta=2$ :
$y = e^{0x}(c_1 \cos 2x + c_2 \sin 2x)$
$y = c_1 \cos 2x + c_2 \sin 2x$

Derive the solution form for a second-order linear homogeneous differential equation when the roots of the auxiliary equation are complex conjugates ( $m = \alpha \pm i\beta$ ).

Let the roots be $m_1 = \alpha + i\beta$ and $m_2 = \alpha - i\beta$ .

From the general solution for distinct roots:
$y = A e^{(\alpha + i\beta)x} + B e^{(\alpha - i\beta)x}$
$y = e^{\alpha x} [ A e^{i\beta x} + B e^{-i\beta x} ]$

Using Euler's formula ( $e^{i\theta} = \cos\theta + i\sin\theta$ ):
$y = e^{\alpha x} [ A(\cos \beta x + i \sin \beta x) + B(\cos \beta x - i \sin \beta x) ]$

Grouping terms:
$y = e^{\alpha x} [ (A+B)\cos \beta x + i(A-B)\sin \beta x ]$

Let $c_1 = A+B$ and $c_2 = i(A-B)$ be arbitrary constants.

The general solution becomes:
$y = e^{\alpha x} (c_1 \cos \beta x + c_2 \sin \beta x)$

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

Step 1: Auxiliary Equation
$m^3 - 6m^2 + 11m - 6 = 0$

Step 2: Find Roots
By inspection, put $m=1$ : $1 - 6 + 11 - 6 = 0$ . So, $(m-1)$ is a factor.
Divide the polynomial by $(m-1)$ to get $m^2 - 5m + 6 = 0$ .
Solving the quadratic: $(m-2)(m-3) = 0$ .

The roots are $m = 1, 2, 3$ (Real and distinct).

Step 3: Solution
$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$

Solve the following differential equation: $(D^4 - 1)y = 0$

Step 1: Auxiliary Equation
$m^4 - 1 = 0$

Step 2: Factorize
$(m^2 - 1)(m^2 + 1) = 0$
$(m - 1)(m + 1)(m^2 + 1) = 0$

Step 3: Determine Roots
From $m^2 - 1 = 0 \implies m = \pm 1$ .
From $m^2 + 1 = 0 \implies m = \pm i$ .
Roots are: $1, -1, 0 \pm i$ .

Step 4: General Solution
Combine real distinct part and complex part:
$y = c_1 e^x + c_2 e^{-x} + e^{0x}(c_3 \cos x + c_4 \sin x)$
$y = c_1 e^x + c_2 e^{-x} + c_3 \cos x + c_4 \sin x$

Solve the Initial Value Problem: $y'' - 3y' + 2y = 0, \quad y(0)=0, \, y'(0)=1$

Step 1: General Solution
A.E.: $m^2 - 3m + 2 = 0 \implies (m-1)(m-2) = 0$ .
Roots: $m_1 = 1, m_2 = 2$ .
$y(x) = c_1 e^x + c_2 e^{2x}$

Step 2: Apply Conditions
Derivative: $y'(x) = c_1 e^x + 2c_2 e^{2x}$ .

Condition 1: $y(0) = 0$
$0 = c_1 e^0 + c_2 e^0 \implies c_1 + c_2 = 0 \implies c_1 = -c_2$

Condition 2: $y'(0) = 1$
$1 = c_1 e^0 + 2c_2 e^0 \implies c_1 + 2c_2 = 1$

Step 3: Solve for Constants
Substitute $c_1 = -c_2$ into the second equation:
$-c_2 + 2c_2 = 1 \implies c_2 = 1$
Then $c_1 = -1$ .

Step 4: Final Solution
$y = -e^x + e^{2x}$

Solve the differential equation where the auxiliary equation has roots $m = 2, 2, 2$ .

The auxiliary equation has a real root $m=2$ repeated three times (multiplicity 3).

Rule for Repeated Roots:
If a root $m$ is repeated $k$ times, the corresponding part of the solution is:
$(c_1 + c_2 x + c_3 x^2 + \dots + c_k x^{k-1})e^{mx}$

Solution:
Here, $k=3$ and $m=2$ .
$y = (c_1 + c_2 x + c_3 x^2)e^{2x}$

Solve: $(D^2 - 2D + 5)^2 y = 0$

Step 1: Solve the quadratic factor
Consider $m^2 - 2m + 5 = 0$ .
Using quadratic formula:
$m = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i$

Step 2: Account for multiplicity
The factor is squared, so the roots $1 \pm 2i$ are repeated twice.
Roots: $1 \pm 2i, 1 \pm 2i$ .

Step 3: General Solution
For repeated complex roots $\alpha \pm i\beta$ (twice), the solution is:
$y = e^{\alpha x} [(c_1 + c_2 x)\cos \beta x + (c_3 + c_4 x)\sin \beta x]$

Substituting $\alpha = 1, \beta = 2$ :
$y = e^x [(c_1 + c_2 x)\cos 2x + (c_3 + c_4 x)\sin 2x]$

Find the differential equation whose general solution is $y = A e^{3x} + B e^{-3x}$

We need to eliminate the arbitrary constants $A$ and $B$ .

Given:
$y = A e^{3x} + B e^{-3x} \quad \dots (1)$

Differentiate w.r.t $x$ :
$y' = 3A e^{3x} - 3B e^{-3x}$

Differentiate again:
$y'' = 9A e^{3x} + 9B e^{-3x}$
$y'' = 9(A e^{3x} + B e^{-3x})$

From equation (1), substitute $y$ back in:
$y'' = 9y$
$y'' - 9y = 0$

Or in operator notation:
$(D^2 - 9)y = 0$

Solve the differential equation: $\frac{d^3y}{dx^3} + y = 0$

Step 1: Auxiliary Equation
$m^3 + 1 = 0$
Using $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ :
$(m + 1)(m^2 - m + 1) = 0$

Step 2: Roots

$m + 1 = 0 \implies m = -1$ .
$m^2 - m + 1 = 0 \implies m = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = \frac{1}{2} \pm i\frac{\sqrt{3}}{2}$ .

Step 3: General Solution
Combine the real root part and the complex root part:
$y = c_1 e^{-x} + e^{x/2} \left( c_2 \cos \frac{\sqrt{3}}{2}x + c_3 \sin \frac{\sqrt{3}}{2}x \right)$

Define the Fundamental Set of Solutions for an $n$ -th order linear differential equation.

For an $n$ -th order homogeneous linear differential equation on an interval $I$ , a set of $n$ solutions $\{y_1, y_2, \dots, y_n\}$ is called a Fundamental Set of Solutions if:

Each $y_i$ is a solution to the differential equation.
The set of functions is Linearly Independent on the interval $I$ (i.e., their Wronskian $W \neq 0$ ).

If such a set exists, the general solution is given by $y = c_1 y_1 + \dots + c_n y_n$ .

Solve: $(D^2 + 16)y = 0$ given that when $x=0, y=0$ and when $x=\frac{\pi}{2}, y=-2$ .

Step 1: General Solution
A.E.: $m^2 + 16 = 0 \implies m = \pm 4i$ .
Solution: $y = c_1 \cos 4x + c_2 \sin 4x$ .

Step 2: Apply Boundary Conditions
At $x=0, y=0$ :
$0 = c_1 \cos(0) + c_2 \sin(0) \implies c_1(1) + 0 = 0 \implies c_1 = 0$
So, $y = c_2 \sin 4x$ .

At $x = \frac{\pi}{2}, y = -2$ :
$-2 = c_2 \sin(4 \cdot \frac{\pi}{2})$
$-2 = c_2 \sin(2\pi)$
Since $\sin(2\pi) = 0$ , we get $-2 = 0$ , which is a contradiction.

Conclusion:
Usually, such problems yield a constant. However, in this specific boundary value problem, no solution exists satisfying both conditions.

Solve the 4th order DE: $y^{(4)} + 2y'' + y = 0$

Step 1: Operator Form and A.E.
$(D^4 + 2D^2 + 1)y = 0$
A.E.: $m^4 + 2m^2 + 1 = 0$

Step 2: Factorize
This is a perfect square quadratic in $m^2$ :
$(m^2 + 1)^2 = 0$

Step 3: Roots
$m^2 + 1 = 0$ is repeated twice.
$m = \pm i$ (repeated twice).
Roots: $0 \pm i, 0 \pm i$ .

Step 4: Solution
For repeated complex roots:
$y = e^{0x} [(c_1 + c_2 x)\cos x + (c_3 + c_4 x)\sin x]$
$y = (c_1 + c_2 x)\cos x + (c_3 + c_4 x)\sin x$

Unit1 Unit3