Question

Let \( \varphi : (0, \infty) \to \mathbb{R} \) be the solution of the differential equation \[ x^2 \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + y = 6x \ln x, \] satisfying \( \varphi(1) = -3 \) and \( \varphi(e) = 0 \). Then, the value of \( \varphi'(1) \) is equal to ............ (rounded off to two decimal places).

Hint:

For solving Cauchy-Euler equations, assume a solution of the form \( y(x) = C_1 x + C_2 x \ln x \) for the homogeneous part. Then use undetermined coefficients for the non-homogeneous part.

Answer 1

Step 1: Solve the homogeneous equation.
The corresponding homogeneous equation is: \[ x^2 \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + y = 0. \] This is a Cauchy-Euler equation, and the solution is of the form: \[ y_h(x) = C_1 x^r + C_2 x^s, \] where \( r \) and \( s \) are the roots of the characteristic equation: \[ r(r - 1) - r + 1 = 0 \implies r^2 = 0 \implies r = 1. \] Thus, the general solution for the homogeneous equation is: \[ y_h(x) = C_1 x + C_2 x \ln x. \] Step 2: Solve the non-homogeneous equation.
Now, solve the non-homogeneous equation using the method of undetermined coefficients. Assume a particular solution of the form: \[ y_p(x) = A x \ln x. \] Substitute into the non-homogeneous equation: \[ x^2 \frac{d^2}{dx^2} \left( A x \ln x \right) - x \frac{d}{dx} \left( A x \ln x \right) + A x \ln x = 6x \ln x. \] After computing the derivatives and solving for \( A \), we find: \[ A = 2. \] Thus, the particular solution is: \[ y_p(x) = 2x \ln x. \] Step 3: General solution.
The general solution is the sum of the homogeneous and particular solutions: \[ y(x) = C_1 x + C_2 x \ln x + 2x \ln x. \] Step 4: Apply the initial conditions.
Use \( \varphi(1) = -3 \) and \( \varphi(e) = 0 \) to find \( C_1 \) and \( C_2 \). At \( x = 1 \): \[ -3 = C_1 + C_2 \cdot 0 + 2 \cdot 0 \implies C_1 = -3. \] At \( x = e \): \[ 0 = -3e + C_2 e \ln e + 2e \ln e \implies -3e + C_2 e + 2e = 0. \] Solving for \( C_2 \), we get: \[ C_2 = 1. \] Step 5: Compute \( \varphi'(x) \).
Now, differentiate \( \varphi(x) = -3x + x \ln x + 2x \ln x \): \[ \varphi'(x) = -3 + \ln x + 2 \ln x + 3 = 3 \ln x. \] Thus, \( \varphi'(1) = 0 \). Final Answer: \[ \boxed{0}. \]