Question

For 𝛼 ∈ (−2𝜋, 0), consider the differential equation
$x^{2} \frac{d^{2} y}{d x^{2}} + α x \frac{d y}{d x} + y = 0 for x > 0 .$
Let 𝐷 be the set of all 𝛼 ∈ (−2𝜋, 0) for which all corresponding real solutions
to the above differential equation approach zero as $𝑥 \to 0^{+}$ . Then, the number of
elements in 𝐷 ∩ ℤ equals _____________

Answer: $\underset{_____________}{6}$

Solution :-

Given that $x^{2} \frac{d^{2} y}{d x^{2}} + α x \frac{d y}{d x} + y = 0$ this is a 2nd order homogeneous linear differential equation with variable coefficient and we know that general solution of heigher order homogeneous linear differential equation is $y = C . F . (complementary function) .$

Let $x = e^{z} ⟹ z = log x \dots \dots (1)$

Now, Auxiliary Equation of given differential equation is

\begin{array}{crcl} D_{1} (D_{1} - 1) + α D_{1} + 1 & = & 0 \\ ⟹ & D_{1}^{2} - D_{1} + α D_{1} + 1 & = & 0 \\ ⟹ & D_{1}^{2} + (α - 1) D_{1} + 1 & = & 0 \\ ⟹ & D_{1} & = & \frac{- (α - 1) \pm \sqrt{{(α - 1)}^{2} - 4}}{2} \\ ∴ & D_{1} & = & \frac{(1 - α) \pm \sqrt{{(α - 1)}^{2} - 4}}{2} \end{array}

Given that 𝐷 be the set of all 𝛼 ∈ (−2𝜋, 0) and we are to find the number of elements in 𝐷 ∩ ℤ such that the solution of the differential equation approaches zero as x → 0⁺, means we can talk about only integer value of 𝛼, which are in (−2𝜋, 0). Interval (−2𝜋, 0) has only six integers which are {-1, -2, -3, -4, -5, -6}.
So, 𝛼 = -1, -2, -3, -4, -5, -6 only.

Now, if 𝛼 = -1 then

\begin{array}{llllllllll} D_{1} & = & \frac{(1 - (- 1)) \pm \sqrt{{((- 1) - 1)}^{2} - 4}}{2} \\ = & \frac{2 \pm \sqrt{{(- 2)}^{2} - 4}}{2} \\ = & \frac{2 \pm 0}{2} = 1, 1 \end{array}

Roots of Auxiliary equation are real and repeated.

\begin{array}{rcl} Hence, C . F . & = & c_{1} e^{z} + c_{2} z e^{z} \\ = & c_{1} x + c_{2} (log x) \cdot x (from eq(1)) \end{array}

So, the general solution is $y (x) = c_{1} x + c_{2} (log x) \cdot x .$
Here, we can see as x → 0⁺, y(x) → 0.

Now, if 𝛼 = -2 then

\begin{array}{llllllllll} D_{1} & = & \frac{(1 - (- 2)) \pm \sqrt{{((- 2) - 1)}^{2} - 4}}{2} \\ = & \frac{3 \pm \sqrt{{(- 3)}^{2} - 4}}{2} \\ = & \frac{3 \pm \sqrt{5}}{2} \\ = & \frac{3}{2} \pm \frac{\sqrt{5}}{2} \end{array}

Here, in this case, the roots of the Auxiliary equation are irrational.

\begin{array}{rcl} Hence, C . F . & = & e^{3 z} (c_{1} cosh (\frac{\sqrt{5}}{2} z) + c_{2} sinh (\frac{\sqrt{5}}{2} z)) \\ = & {(e^{z})}^{3} (c_{1} cosh (\frac{\sqrt{5}}{2} log x) + c_{2} sinh (\frac{\sqrt{5}}{2} log x)) \\ = & x^{3} (c_{1} cosh (\frac{\sqrt{5}}{2} log x) + c_{2} sinh (\frac{\sqrt{5}}{2} log x)) \end{array}

So, the general solution is $y (x) = x^{3} (c_{1} cosh (\frac{\sqrt{5}}{2} log x) + c_{2} sinh (\frac{\sqrt{5}}{2} log x)) .$
Here, we can see as x → 0⁺, y(x) → 0.

Now, we can observe that if 𝛼 = -3, -4, -5, -6 then $D_{1}$ gives irrational roots, so forall $α \in {- 3, - 4, - 5, - 6}$ , the solutions $y (x)$ are the same as 𝛼 = -2 (above) the only difference is the power of x and scalar which multiplied with $log x$ .
Hence, forall $α \in {- 3, - 4, - 5, - 6}$ the solutions $y (x)$ approaches 0 as $𝑥 \to 0^{+} .$

Therefore, the solutions of the given differential equation approaches zero forall $α \in {- 1 - 2 - 3, - 4, - 5, - 6}$ . Hence, all the elements of 𝐷 ∩ ℤ will be counted, and the number of elements in 𝐷 ∩ ℤ is equal to 6.

So, the correct answer is $\to$ $\underset{_____________}{6}$

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