Question

Consider the following statements:
A. $(1 + e^x y + x e^x y) dx + (x e^x + 2) dy = 0$ is an exact differential equation.
B. The particular solution of $(D^2 - D - 2)y = e^{-x}$ is $-\frac{1}{3} x e^{-x}$. C. The particular solution of $(D^2 + 4)y = \sin^2 x$ is $-\frac{x}{8} \sin 2x$. D. The functions $\phi_1(x) = x^2$ and $\phi_2(x) = x |x|$ are linearly independent for $-\infty < x < \infty$.
Choose the correct answer from the options given below:

• A. A, C Only
• B. B, C Only
• C. A, B, D Only
• D. A, B, C Only

Hint:

The functions $x^2$ and $x|x|$ have Wronskian equal to $0$ everywhere, yet they are linearly independent on $\mathbb{R}$! This demonstrates that $W = 0$ does not imply linear dependence for non-analytic functions.

Answer 1

The Correct Option is C

Step 1 : Concept:
This question tests test for exact differential equations, finding particular integrals for non-homogeneous linear differential equations, and linear independence of functions.

Step 2 : Key Formulas and Approach:
1. Exactness condition for $M dx + N dy = 0$: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] 2. Particular integral for $P(D)y = e^{ax}$ when $P(a) = 0$: \[ y_p = \frac{1}{P(D)} e^{ax} = \frac{x}{P'(a)} e^{ax} \] 3. Functions $\phi_1, \phi_2$ are linearly independent if $c_1 \phi_1(x) + c_2 \phi_2(x) = 0$ for all $x \implies c_1 = c_2 = 0$.

Step 3 : Step-by-step Explanation:

• Statement A:
Here $M(x,y) = 1 + e^x y + x e^x y$ and $N(x,y) = x e^x + 2$. \[ \frac{\partial M}{\partial y} = e^x + x e^x \] \[ \frac{\partial N}{\partial x} = 1 \cdot e^x + x e^x = e^x + x e^x \] Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact. Statement A is correct.

• Statement B:
For $(D^2 - D - 2)y = e^{-x}$, $P(D) = D^2 - D - 2$.
$P(-1) = (-1)^2 - (-1) - 2 = 0$ (resonance).
$P'(D) = 2D - 1 \implies P'(-1) = 2(-1) - 1 = -3$.
Using the shift formula: \[ y_p = \frac{x}{P'(-1)} e^{-x} = \frac{x}{-3} e^{-x} = -\frac{1}{3} x e^{-x} \] Statement B is correct.

• Statement C:
For $(D^2 + 4)y = \sin^2 x = \frac{1 - \cos 2x}{2}$: \[ y_p = \frac{1}{D^2 + 4} \left(\frac{1}{2}\right) - \frac{1}{D^2 + 4}\left(\frac{\cos 2x}{2}\right) = \frac{1}{8} - \frac{1}{2}\left(\frac{x}{4} \sin 2x\right) = \frac{1}{8} - \frac{x}{8} \sin 2x \] The statement omits the constant term $\frac{1}{8}$, making it incomplete/incorrect as the particular solution. Statement C is incorrect.

• Statement D:
Consider $c_1 x^2 + c_2 x|x| = 0$ for all $x \in (-\infty, \infty)$.
For $x = 1$: $c_1(1)^2 + c_2(1)(1) = 0 \implies c_1 + c_2 = 0$.
For $x = -1$: $c_1(-1)^2 + c_2(-1)|-1| = 0 \implies c_1 - c_2 = 0$.
Solving these simultaneously yields $c_1 = 0$ and $c_2 = 0$.
Hence, $\phi_1(x) = x^2$ and $\phi_2(x) = x|x|$ are linearly independent on $(-\infty, \infty)$. Statement D is correct.

Step 4 : Final Answer:
Statements A, B, and D are correct. Therefore, option (C) is the correct answer.