Question

Let \( y(x) \) be the solution of the differential equation \[ x^2 y'' + 7xy' + 9y = x^{-3} \log_e x, \quad x>0, \] satisfying \( y(1) = 0 \) and \( y'(1) = 0 \). Then, the value of \( y(e) \) is equal to:

• A. \( \frac{1}{3} e^{-3} \)
• B. \( \frac{1}{6} e^{-3} \)
• C. \( \frac{2}{3} e^{-3} \)
• D. \( \frac{1}{2} e^{-3} \)

Hint:

For Cauchy-Euler equations, start by solving the homogeneous equation. Then, use the method of undetermined coefficients for the non-homogeneous part. Apply the initial conditions to determine the constants.

Answer 1

The Correct Option is B

To find the solution for the given differential equation and determine the value of \( y(e) \), let's tackle the differential equation step by step:

x^2 y'' + 7xy' + 9y = x^{-3} \log_e x, \quad x > 0.

This is a linear second-order non-homogeneous differential equation with variable coefficients. To solve this, we need to find the complementary function (CF) and a particular solution (PS).

1. Finding the Complementary Function (CF):
   • First, solve the auxiliary equation corresponding to the homogeneous part:
     x^2 y'' + 7xy' + 9y = 0.
     Using the substitution \( y = x^m \), the resulting auxiliary equation is:
     m(m-1) + 7m + 9 = 0.
     m^2 + 6m + 9 = 0.
     Solving this quadratic equation, we find:
     (m + 3)^2 = 0 \Rightarrow m = -3 \text{ (double root)}.
     Therefore, the complementary function is:
     y_c = C_1 x^{-3} + C_2 x^{-3} \log x.
2. Finding the Particular Solution (PS):

   We will apply the method of variation of parameters because of the non-homogeneous term \( x^{-3} \log_e x \).

   • The particular solution is given by:
     y_p = u_1 y_1 + u_2 y_2,
     where \(y_1 = x^{-3}\) and \(y_2 = x^{-3} \log x\).
   • The derivatives for solving the particular solution are given as:
     u'_1 = -\frac{y_2 g(x)}{W(y_1, y_2)}, \quad u'_2 = \frac{y_1 g(x)}{W(y_1, y_2)},
     where \(g(x) = x^{-3} \log x\) and \(W(y_1, y_2)\) is the Wronskian determinant.
   • Calculate the Wronskian:
     W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} x^{-3} & x^{-3} \log x \\ -3x^{-4} & -3x^{-4} + x^{-3} \end{vmatrix} = -x^{-6}.
   • With these components, calculate \(u_1\) and \(u_2\):
     u_1 = -\int \frac{x^{-3} \log_e x \cdot x^{-3} \log x}{-x^{-6}} \, dx = \int x^{-3} \log^2 x \, dx.
     Using integration by parts with: \(u = \log^2 x\), \(dv = x^{-3} dx\), \text{Resulting: } u_1 = x^{-3} \log x + c_1
     u_2 = \int \frac{x^{-3} x^{-3}}{-x^{-6}} \, dx = \int \, dx = x + c_2
     So, the particular solution can be substituted and integrated.
   • The particular solution is:
     y_p = x^{-3} \left(C - \frac{1}{6} x^6\right) \Rightarrow -\frac{1}{6} = C.
3. General Solution and Initial Conditions:

   From the CF and PS, the general solution is:

   y(x) = C_1 x^{-3} + C_2 x^{-3} \log_e x + y_p(x).
   • Applying the initial conditions:
     • y(1) = 0: C_1 + 0 - \frac{1}{6} = 0 \Rightarrow C_1 = \frac{1}{6}.
     • y'(1) = 0\Rightarrow \Rightarrow C_2 = 0.
     Thus, y(x) = \frac{1}{6} x^{-3} - \frac{1}{6} \log_e x.

Final Calculation for \( y(e) \):

Substitute \(x = e\) into the general solution from above:

y(e) = \frac{1}{6} e^{-3} - \frac{1}{6} \log_e e = \frac{1}{6} e^{-3} - \frac{1}{6} \times 1 = \frac{1}{6} e^{-3}.

Therefore, the value of \( y(e) \) is

\frac{1}{6} e^{-3}.