Differential Equations

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2024 Q.64)** **1. Governing Concept & Definition**: - **Topic**: Second-Order Cauchy-Euler Differential Equations - An equation of the form $x^2 y'' + 4y = lpha x y'$ can be solved by substituting $y(x) = x^2 \ln x$ and its derivatives. **2. Detailed Step-by-Step Calculation**: - **Given Function**: $y(x) = x^2 \ln x$ - **First Derivative**: $y'(x) = 2x \ln x + x^2 \left(rac{1}{x} ight) = 2x \ln x + x$ - **Second Derivative**: $y''(x) = 2 \ln x + 2x \left(rac{1}{x} ight) + 1 = 2 \ln x + 3$ - **Substituting into ODE**: $x^2(2 \ln x + 3) + 4(x^2 \ln x) = lpha x (2x \ln x + x)$ $2x^2 \ln x + 3x^2 + 4x^2 \ln x = lpha (2x^2 \ln x + x^2)$ $(6x^2 \ln x + 3x^2) = lpha (2x^2 \ln x + x^2)$ $3(2x^2 \ln x + x^2) = lpha (2x^2 \ln x + x^2) \implies lpha = 3$ **3. Final Answer**: - **Calculated Value**: **3** - **Official GATE Accepted Range**: **3 to 3**