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Unit3 - Subjective Questions

MTH166 • Practice Questions with Detailed Answers

Solve the differential equation $(D^2 - 5D + 6)y = e^{4x}$ using the operator method.

1. Complementary Function (C.F.):
The auxiliary equation is $m^2 - 5m + 6 = 0$ .
$(m-2)(m-3) = 0 \implies m = 2, 3$
$C.F. = c_1 e^{2x} + c_2 e^{3x}$

2. Particular Integral (P.I.):
$P.I. = \frac{1}{D^2 - 5D + 6} e^{4x}$
Replace $D$ with $a=4$ :
$P.I. = \frac{1}{4^2 - 5(4) + 6} e^{4x} = \frac{1}{16 - 20 + 6} e^{4x} = \frac{1}{2} e^{4x}$

3. General Solution:
$y = C.F. + P.I.$
$y = c_1 e^{2x} + c_2 e^{3x} + \frac{1}{2} e^{4x}$

Solve the differential equation $(D^2 + 4)y = \sin 3x$ .

1. Complementary Function (C.F.):
Auxiliary equation: $m^2 + 4 = 0 \implies m = \pm 2i$ .
$C.F. = c_1 \cos 2x + c_2 \sin 2x$

2. Particular Integral (P.I.):
$P.I. = \frac{1}{D^2 + 4} \sin 3x$
Replace $D^2$ with $-a^2 = -3^2 = -9$ :
$P.I. = \frac{1}{-9 + 4} \sin 3x = \frac{1}{-5} \sin 3x$

3. General Solution:
$y = c_1 \cos 2x + c_2 \sin 2x - \frac{1}{5} \sin 3x$

Explain the method of finding the Particular Integral when the non-homogeneous part is of the form $x^m$ (a polynomial). Solve $(D^2 + D)y = x^2$ .

Explanation:
When $X = x^m$ , we expand the operator $[f(D)]^{-1}$ in ascending powers of $D$ using Binomial expansion (e.g., $(1+D)^{-1}$ or $(1-D)^{-1}$ ) and operate on $x^m$ . Terms in $D$ higher than order $m$ will yield zero.

Solution for $(D^2 + D)y = x^2$ :

1. C.F.:
$m(m+1)=0 \implies m=0, -1$ .
$C.F. = c_1 + c_2 e^{-x}$

2. P.I.:
$P.I. = \frac{1}{D(1+D)} x^2 = \frac{1}{D} (1+D)^{-1} x^2$
Expand $(1+D)^{-1} = 1 - D + D^2 - \dots$
$P.I. = \frac{1}{D} [ (1 - D + D^2) x^2 ]$
$P.I. = \frac{1}{D} [ x^2 - 2x + 2 ]$
Integrate each term (since $\frac{1}{D}$ is integration):
$P.I. = \frac{x^3}{3} - x^2 + 2x$

3. General Solution:
$y = c_1 + c_2 e^{-x} + \frac{x^3}{3} - x^2 + 2x$

Solve $(D^2 - 4D + 4)y = e^{2x}$ using the operator method (Case of Failure).

1. Complementary Function (C.F.):
Auxiliary equation: $m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0 \implies m = 2, 2$ (Repeated roots).
$C.F. = (c_1 + c_2 x)e^{2x}$

2. Particular Integral (P.I.):
$P.I. = \frac{1}{(D-2)^2} e^{2x}$
If we put $D=2$ , the denominator becomes zero (Failure Case).
Apply formula: $\frac{1}{(D-a)^n} e^{ax} = \frac{x^n}{n!} e^{ax}$ .
Here $n=2, a=2$ .
$P.I. = \frac{x^2}{2!} e^{2x} = \frac{x^2}{2} e^{2x}$

3. General Solution:
$y = (c_1 + c_2 x)e^{2x} + \frac{x^2}{2} e^{2x}$

Using the method of Variation of Parameters, solve $\frac{d^2y}{dx^2} + y = \sec x$ .

1. Complementary Function:
$m^2 + 1 = 0 \implies m = \pm i$ .
$y_c = c_1 \cos x + c_2 \sin x$
Here, $y_1 = \cos x$ and $y_2 = \sin x$ .

2. Wronskian ( $W$ ):
$W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1$

3. Particular Integral ( $y_p = u y_1 + v y_2$ ):
$u = -\int \frac{y_2 X}{W} dx = -\int \frac{\sin x \cdot \sec x}{1} dx = -\int \tan x \, dx = -\log|\sec x| = \log|\cos x|$
$v = \int \frac{y_1 X}{W} dx = \int \frac{\cos x \cdot \sec x}{1} dx = \int 1 \, dx = x$

$y_p = (\log|\cos x|) \cos x + x \sin x$

4. General Solution:
$y = c_1 \cos x + c_2 \sin x + \cos x \log|\cos x| + x \sin x$

Explain the substitution required to transform an Euler-Cauchy equation $x^n \frac{d^n y}{dx^n} + \dots + ay = X$ into a linear differential equation with constant coefficients.

To solve the Euler-Cauchy equation (homogeneous linear equation with variable coefficients), use the substitution:

Let $x = e^z$ or $z = \log x$ .
This implies $\frac{dy}{dx} = \frac{1}{x} \frac{dy}{dz}$ and $\frac{d^2y}{dx^2} = \frac{1}{x^2}(\frac{d^2y}{dz^2} - \frac{dy}{dz})$ .

Define the operator $\theta = \frac{d}{dz}$ . The transformations are:

$x \frac{dy}{dx} \rightarrow \theta y$
$x^2 \frac{d^2y}{dx^2} \rightarrow \theta(\theta - 1)y$
$x^3 \frac{d^3y}{dx^3} \rightarrow \theta(\theta - 1)(\theta - 2)y$

Substituting these into the original equation results in a differential equation with constant coefficients in terms of independent variable $z$ , which can be solved using standard methods. Finally, replace $z$ with $\log x$ in the solution.

Solve the Euler-Cauchy equation: $x^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} + y = \log x$ .

1. Transformation:
Put $x = e^z, z = \log x$ . Let $\theta = \frac{d}{dz}$ .
$x^2 y'' \to \theta(\theta-1)y$ , $x y' \to \theta y$ .
Equation becomes: $[\theta(\theta - 1) - \theta + 1]y = z$
$(\theta^2 - 2\theta + 1)y = z$

2. Complementary Function (C.F.):
Auxiliary eq: $(\theta - 1)^2 = 0 \implies m = 1, 1$ .
$C.F. = (c_1 + c_2 z)e^z$

3. Particular Integral (P.I.):
$P.I. = \frac{1}{(\theta - 1)^2} z = (1 - \theta)^{-2} z$
Expand: $(1 + 2\theta + \dots) z = z + 2(1) = z + 2$ .

4. General Solution in terms of $x$ :
$y = (c_1 + c_2 z)e^z + z + 2$
Substitute $z = \log x, e^z = x$ :
$y = x(c_1 + c_2 \log x) + \log x + 2$

Solve $(D^2 - 2D + 1)y = x e^x \sin x$ using the shift theorem.

1. C.F.:
$(m-1)^2 = 0 \implies m=1,1$ . $C.F. = (c_1 + c_2 x)e^x$ .

2. P.I.:
$P.I. = \frac{1}{(D-1)^2} (e^x \cdot x \sin x)$
Using shift theorem $\frac{1}{f(D)} e^{ax} V = e^{ax} \frac{1}{f(D+a)} V$ :
$P.I. = e^x \frac{1}{((D+1)-1)^2} (x \sin x) = e^x \frac{1}{D^2} (x \sin x)$
Integrate $x \sin x$ twice.
First integration ( $\int x \sin x dx$ ): $-x \cos x + \sin x$ .
Second integration ( $\int (-x \cos x + \sin x) dx$ ):
$= - [x \sin x - \int \sin x dx] - \cos x$
$= - [x \sin x + \cos x] - \cos x = -x \sin x - 2\cos x$

$P.I. = -e^x (x \sin x + 2\cos x)$

3. General Solution:
$y = (c_1 + c_2 x)e^x - e^x(x \sin x + 2\cos x)$

Solve the simultaneous differential equations: $\frac{dx}{dt} + y = \sin t$ and $x + \frac{dy}{dt} = \cos t$ .

Write in operator form ( $D = d/dt$ ):
1) $Dx + y = \sin t$
2) $x + Dy = \cos t$

Multiply (1) by $D$ and (2) by 1:
3) $D^2x + Dy = D(\sin t) = \cos t$
4) $x + Dy = \cos t$

Subtract (4) from (3):
$(D^2 - 1)x = 0$
Auxiliary eq: $m^2 - 1 = 0 \implies m = \pm 1$ .
$x(t) = c_1 e^t + c_2 e^{-t}$

Now find $y$ . From (1), $y = \sin t - Dx$ .
$Dx = c_1 e^t - c_2 e^{-t}$
$y = \sin t - (c_1 e^t - c_2 e^{-t}) = \sin t - c_1 e^t + c_2 e^{-t}$

Solution:
$x = c_1 e^t + c_2 e^{-t}$
$y = -c_1 e^t + c_2 e^{-t} + \sin t$

Solve $(D^2 + a^2)y = \cos ax$ (Resonance Case).

1. Complementary Function:
$m^2 + a^2 = 0 \implies m = \pm ai$ .
$C.F. = c_1 \cos ax + c_2 \sin ax$

2. Particular Integral:
$P.I. = \frac{1}{D^2 + a^2} \cos ax$
Since replacing $D^2$ with $-a^2$ gives 0, use the formula $\frac{1}{D^2+a^2} \cos ax = \frac{x}{2a} \sin ax$ .
$P.I. = \frac{x}{2a} \sin ax$

3. General Solution:
$y = c_1 \cos ax + c_2 \sin ax + \frac{x}{2a} \sin ax$

Using the Method of Undetermined Coefficients, determine the form of the particular solution for $y'' - 4y' + 4y = 2e^{2x} + x^2$ .

First, find the Complementary Function (C.F.):
$(m-2)^2 = 0 \implies y_c = c_1 e^{2x} + c_2 x e^{2x}$ .

The non-homogeneous term has two parts: $2e^{2x}$ and $x^2$ .

For $2e^{2x}$ :
Standard guess is $A e^{2x}$ . However, $e^{2x}$ and $xe^{2x}$ are already in the C.F. (roots repeated twice).
We must multiply by $x^2$ .
$y_{p1} = A x^2 e^{2x}$ .

For $x^2$ :
Standard guess for a quadratic polynomial is $B x^2 + C x + E$ .
None of these terms appear in the C.F.

Combined Form of $y_p$ :
$y_p = A x^2 e^{2x} + B x^2 + C x + E$

Solve using the Method of Variation of Parameters: $y'' - 2y' + y = e^x \log x$ .

1. C.F.:
$(m-1)^2 = 0 \implies y_1 = e^x, y_2 = x e^x$ .
$y_c = c_1 e^x + c_2 x e^x$ .

2. Wronskian:
$W = (e^x)(e^x + xe^x) - (xe^x)(e^x) = e^{2x} + xe^{2x} - xe^{2x} = e^{2x}$ .

3. Formula:
$u = -\int \frac{y_2 X}{W} dx = -\int \frac{xe^x (e^x \log x)}{e^{2x}} dx = -\int x \log x dx$ .
Integration by parts: $-[\frac{x^2}{2}\log x - \int \frac{x^2}{2} \frac{1}{x} dx] = -\frac{x^2}{2}\log x + \frac{x^2}{4}$ .

$v = \int \frac{y_1 X}{W} dx = \int \frac{e^x (e^x \log x)}{e^{2x}} dx = \int \log x dx = x \log x - x$ .

4. P.I.:
$y_p = u y_1 + v y_2 = (-\frac{x^2}{2}\log x + \frac{x^2}{4})e^x + (x \log x - x)x e^x$
$y_p = e^x [ -\frac{x^2}{2}\log x + \frac{x^2}{4} + x^2\log x - x^2 ]$
$y_p = e^x [ \frac{x^2}{2}\log x - \frac{3x^2}{4} ]$

General Solution: $y = y_c + y_p$ .

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

Auxiliary Equation:
$m^4 - 1 = 0$
$(m^2 - 1)(m^2 + 1) = 0$
$(m-1)(m+1)(m^2+1) = 0$

Roots:
$m = 1, -1$ (Real and Distinct)
$m = \pm i$ (Complex conjugates)

General Solution:
$y = c_1 e^x + c_2 e^{-x} + c_3 \cos x + c_4 \sin x$
Alternatively using hyperbolic functions:
$y = A \cosh x + B \sinh x + c_3 \cos x + c_4 \sin x$

Solve $(D^2 - 4D + 3)y = \sin 3x \cos 2x$ .

1. Pre-processing:
Convert product to sum: $\sin 3x \cos 2x = \frac{1}{2}[\sin(5x) + \sin(x)]$ .
Equation: $(D^2 - 4D + 3)y = \frac{1}{2}\sin 5x + \frac{1}{2}\sin x$ .

2. C.F.:
$(m-3)(m-1)=0 \implies y_c = c_1 e^{3x} + c_2 e^x$ .

3. P.I.:
$P.I. = \frac{1}{2} \frac{1}{D^2-4D+3} \sin 5x + \frac{1}{2} \frac{1}{D^2-4D+3} \sin x$

For $\sin 5x$ ( $D^2 o -25$ ):
$\frac{1}{-25 - 4D + 3} = \frac{1}{-22-4D} = -\frac{1}{2(11+2D)} \cdot \frac{11-2D}{11-2D} = -\frac{11-2D}{2(121-4D^2)}$ .
Sub $D^2=-25$ : Denom $= 2(121 + 100) = 442$ .
Term 1 $= -\frac{11 \sin 5x - 10 \cos 5x}{442}$ .

For $\sin x$ ( $D^2 o -1$ ):
$\frac{1}{-1 - 4D + 3} = \frac{1}{2-4D} = \frac{1}{2(1-2D)} \cdot \frac{1+2D}{1-4D^2}$ .
Sub $D^2=-1$ : Denom $= 2(1+4) = 10$ .
Term 2 $= \frac{1}{20}(\sin x + 2\cos x)$ .

Combine for General Solution.

Find the general solution of the Euler-Cauchy equation $x^2 y'' + 4xy' + 2y = e^x$ .

1. Substitution:
$x = e^z, z = \log x, \theta = d/dz$ .
Eq: $[ heta( heta-1) + 4\theta + 2]y = e^{e^z}$
$( heta^2 + 3\theta + 2)y = e^{e^z}$ .

2. C.F.:
$( heta+1)( heta+2) = 0 \implies m = -1, -2$ .
$y_c = c_1 e^{-z} + c_2 e^{-2z} = c_1 x^{-1} + c_2 x^{-2}$ .

3. P.I.:
$P.I. = \frac{1}{(\theta+1)(\theta+2)} e^{e^z}$ .
Using partial fractions or variation of parameters on the z-domain is complex. However, notice the RHS is not a simple $e^{kz}$ . This often requires Variation of Parameters after transforming to constant coefficients.

Let's apply Var of Param on $y(z)$ : $y_1=e^{-z}, y_2=e^{-2z}$ .
$W = -e^{-3z} - (-2e^{-3z}) = e^{-3z}$ .
$u = -\int \frac{e^{-2z} e^{e^z}}{e^{-3z}} dz = -\int e^z e^{e^z} dz = -e^{e^z}$ .
$v = \int \frac{e^{-z} e^{e^z}}{e^{-3z}} dz = \int e^{2z} e^{e^z} dz$ . Let $e^z=t$ , $dt=e^z dz$ . $\int t e^t dt = te^t - e^t = e^{e^z}(e^z - 1)$ .

$y_p = (-e^{e^z})e^{-z} + e^{e^z}(e^z-1)e^{-2z} = e^{e^z}(-e^{-z} + e^{-z} - e^{-2z}) = -e^{-2z} e^{e^z}$ .
Substitute back $e^z=x$ : $y_p = -\frac{1}{x^2} e^x$ .

General Solution: $y = \frac{c_1}{x} + \frac{c_2}{x^2} - \frac{e^x}{x^2}$ .

Solve $(D^2 + 1)y = x^2 \sin 2x$ .

1. C.F.:
$y_c = c_1 \cos x + c_2 \sin x$ .

2. P.I.:
$P.I. = \frac{1}{D^2+1} \text{Im}(x^2 e^{i2x})$ (Complex variable method) OR perform $Im(\frac{1}{D^2+1} x^2 e^{i2x})$ .
Alternatively, standard formula involves repeated parts.

Using Shift Theorem on $x^2 e^{2ix}$ :
$e^{2ix} \frac{1}{(D+2i)^2+1} x^2 = e^{2ix} \frac{1}{D^2 + 4iD - 3} x^2$ .
Factor out $-3$ : $\frac{e^{2ix}}{-3} [1 - \frac{D^2+4iD}{3}]^{-1} x^2$ .
Expand and solve for the imaginary part.

This is a calculation-intensive question suitable for 10 marks testing algebraic manipulation.

Define the Wronskian of two functions $y_1$ and $y_2$ . State the condition for linear independence based on the Wronskian.

Definition:
The Wronskian $W$ of two differentiable functions $y_1(x)$ and $y_2(x)$ is defined as the determinant:
$W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1'$

Condition for Linear Independence:
If $y_1$ and $y_2$ are solutions to a homogeneous linear differential equation on an interval $I$ :

If $W(y_1, y_2) \neq 0$ for all $x \in I$ , the functions are Linearly Independent.
If $W(y_1, y_2) = 0$ for all $x \in I$ , the functions are Linearly Dependent.

Solve the simultaneous equations $\frac{dx}{dt} + 5x - 2y = t$ and $\frac{dy}{dt} + 2x + y = 0$ .

1. Operator Form:
(1) $(D+5)x - 2y = t$
(2) $2x + (D+1)y = 0 \implies x = -\frac{1}{2}(D+1)y$

2. Substitute x into (1):
$(D+5)[-\frac{1}{2}(D+1)y] - 2y = t$
Multiply by -2: $(D+5)(D+1)y + 4y = -2t$
$(D^2 + 6D + 5)y + 4y = -2t$
$(D^2 + 6D + 9)y = -2t$

3. Solve for y:
Aux: $(m+3)^2 = 0 \implies y_c = (c_1 + c_2 t)e^{-3t}$ .
PI: $\frac{1}{(D+3)^2} (-2t) = -2 (3(1+D/3))^{-2} t = -\frac{2}{9} (1 - \frac{2D}{3}) t = -\frac{2}{9}(t - 2/3)$ .
$y = (c_1+c_2 t)e^{-3t} - \frac{2}{9}t + \frac{4}{27}$ .

4. Find x:
Use $x = -\frac{1}{2}(y' + y)$ . Calculate derivative of $y$ and substitute back.

Using the method of undetermined coefficients, solve $(D^2 - 1)y = \cosh x$ .

1. Rewrite RHS: $\cosh x = \frac{e^x + e^{-x}}{2}$ .
Eq: $(D^2 - 1)y = \frac{1}{2}e^x + \frac{1}{2}e^{-x}$ .

2. C.F.:
$m^2 - 1 = 0 \implies m = \pm 1$ .
$y_c = c_1 e^x + c_2 e^{-x}$ .

3. Form of Particular Solution:
Since $e^x$ and $e^{-x}$ are in C.F., multiply guess by $x$ .
$y_p = A x e^x + B x e^{-x}$ .

4. Solve Coefficients:
Calculate $y_p''$ and substitute into $(D^2-1)y_p$ .
Result: $2A e^x - 2B e^{-x} = \frac{1}{2}e^x + \frac{1}{2}e^{-x}$ .
$2A = 1/2 \implies A = 1/4$ .
$-2B = 1/2 \implies B = -1/4$ .

5. General Solution:
$y = c_1 e^x + c_2 e^{-x} + \frac{1}{4} x e^x - \frac{1}{4} x e^{-x} = c_1 e^x + c_2 e^{-x} + \frac{x}{2} \sinh x$ .

Derive the formula for the Particular Integral of $\frac{d^2y}{dx^2} + ay = X$ using the Method of Variation of Parameters.

Consider the equation $y'' + P y' + Q y = X$ . Let the complementary function be $y_c = c_1 y_1 + c_2 y_2$ .
Assume the particular solution is $y_p = u(x)y_1 + v(x)y_2$ .

Differentiating $y_p$ :
$y_p' = u'y_1 + u y_1' + v'y_2 + v y_2'$ .
Assume condition $u'y_1 + v'y_2 = 0$ (1).
Then $y_p' = u y_1' + v y_2'$ .
Differentiate again: $y_p'' = u'y_1' + u y_1'' + v'y_2' + v y_2''$ .

Substitute into ODE:
$u(y_1''+Py_1'+Qy_1) + v(y_2''+Py_2'+Qy_2) + u'y_1' + v'y_2' = X$ .
Since $y_1, y_2$ are solutions to homogeneous part, the terms in brackets are 0.
We get $u'y_1' + v'y_2' = X$ (2).

Solving system (1) and (2) for $u', v'$ using Cramer's rule:
Denominator is Wronskian $W = y_1 y_2' - y_2 y_1'$ .
$u' = \frac{-y_2 X}{W} \implies u = \int \frac{-y_2 X}{W} dx$
$v' = \frac{y_1 X}{W} \implies v = \int \frac{y_1 X}{W} dx$

Thus, $P.I. = y_1 \int \frac{-y_2 X}{W} dx + y_2 \int \frac{y_1 X}{W} dx$ .

Unit2 Unit4