Question

Let 𝑓: ℝ → ℝ be a solution of the differential equation
$\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + y = 2 e^{x} for x \in ℝ .$
Consider the following statements.
P: If 𝑓(𝑥) > 0 for all 𝑥 ∈ ℝ, then 𝑓′(𝑥) > 0 for all 𝑥 ∈ ℝ.

Q: If 𝑓’(𝑥) > 0 for all 𝑥 ∈ ℝ, then 𝑓(𝑥) > 0 for all 𝑥 ∈ ℝ.

Then, which one of the following holds?

P is true but Q is false
P is false but Q is true
Both P and Q are true
Both P and Q are false

Answer: P is false but Q is true

Solution :-

Given that $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + y = 2 e^{x},$ this is a 2nd order linear differential equation with constant coefficient and we know that general solution of heigher order linear differential equation is

f (x) = C . F . (complementary function) + P . I . (particular integral) .

Auxiliary Equation of given differential equation is

\begin{array}{crcl} D^{2} - 2 D + 1 & = & 0 \\ ⟹ & {(D - 1)}^{2} & = & 0 \\ ∴ & D & = & 1, 1. \end{array}

Roots of Auxiliary equation is real and repeated.
Hence, $C . F . = c_{1} e^{x} + c_{2} x e^{x} .$

Now, $P . I . = \frac{1}{F (D)} \cdot Q (x)$ here $F (D) = {(D - 1)}^{2}$ and $Q (x) = 2 e^{x} .$

\begin{array}{llllllllll} ∴ P . I . & = & \frac{1}{{(D - 1)}^{2}} \cdot 2 e^{x} \\ = & \frac{x}{2 (D - 1)} \cdot 2 e^{x} \\ = & \frac{x^{2}}{1} \cdot e^{x} \\ = & x^{2} e^{x} . \end{array}

So, the general solution is $f (x) = C . F . + P . I . = c_{1} e^{x} + c_{2} x e^{x} + x^{2} e^{x}$

f (x) = (c_{1} + c_{2} x + x^{2}) e^{x} .

\begin{array}{llllllllll} Now, & f^{'} (x) & = & c_{1} e^{x} + c_{2} e^{x} + c_{2} x e^{x} + 2 x e^{x} + x^{2} e^{x} \\ ∴ & f^{'} (x) & = & [(c_{1} + c_{2}) + (c_{2} + 2) x + x^{2}] e^{x} . \end{array}

Case (I) : $f (x) > 0$

This is possible only if $c_{1} + c_{2} x + x^{2} > 0$ and $c_{1} + c_{2} x + x^{2} > 0$ only if coefficient of $x^{2}$ , greater than zero and discriminant(D), less than zero.

Coefficient of $x^{2}$ is 1 > 0. Now, discriminant(D) < 0

\begin{array}{rcl} {(c_{2})}^{2} - 4 \cdot 1 \cdot c_{1} & < & 0 \\ {(c_{2})}^{2} & < & 4 c_{1} . \end{array}

Case (II) : $f^{'} (x) > 0$

This is possible only if $(c_{1} + c_{2}) + (c_{2} + 2) x + x^{2} > 0$ and $(c_{1} + c_{2}) + (c_{2} + 2) x + x^{2} > 0$ only if coefficient of $x^{2}$ , greater than zero and discriminant(D), less than zero.

Coefficient of $x^{2}$ is 1 > 0. Now, discriminant(D) < 0

\begin{array}{rcl} {(c_{2} + 2)}^{2} - 4 \cdot 1 \cdot (c_{1} + c_{2}) & < & 0 \\ {(c_{2})}^{2} + 4 c_{2} + 4 - 4 c_{1} - 4 c_{2} & < & 0 \\ {(c_{2})}^{2} + 4 & < & 4 c_{1} . \end{array}

Now, $f (x) > 0 ⇏ f^{'} (x) > 0 \forall x \in ℝ$ , because of ${(c_{2})}^{2} < 4 c_{1} ⇏ {(c_{2})}^{2} + 4 < 4 c_{1} \forall c_{1}, c_{2} \in ℝ$

But $f^{'} (x) > 0 ⟹ f (x) > 0 \forall x \in ℝ$ , because of ${(c_{2})}^{2} + 4 < 4 c_{1} ⟹ {(c_{2})}^{2} < 4 c_{1} \forall c_{1}, c_{2} \in ℝ .$

So, the correct answer is option-2. $\to$ P is false but Q is true

#Ordinary Differential Equation #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs