Given differential equation: \[ x^2 \frac{d^2y}{dx^2} - 3x \frac{dy}{dx} - 12y = e^{6\ln x} + \ln x \] Let's solve step by step: --- ## **Step 1: Simplify the Equation** First, note that \( e^{6\ln x} = (e^{\ln x})^6 = x^6 \). So the equation becomes: \[ x^2 y'' - 3x y' - 12y = x^6 + \ln x \] --- ## **Step 2: Solve the Homogeneous Equation** The homogeneous equation is: \[ x^2 y'' - 3x y' - 12y = 0 \] This is a Cauchy-Euler (Euler-Cauchy) equation. Substitute \( y = x^m \): \[ y' = m x^{m-1} \\ y'' = m(m-1) x^{m-2} \] Substitute into the equation: \[ x^2[m(m-1)x^{m-2}] - 3x[mx^{m-1}] - 12x^m = 0 \\ m(m-1)x^m - 3m x^m - 12x^m = 0 \\ [m(m-1) - 3m - 12] x^m = 0 \\ [m^2 - m - 3m - 12] x^m = 0 \\ [m^2 - 4m - 12] x^m = 0 \] So, the characteristic equation is: \[ m^2 - 4m - 12 = 0 \] Solve for \( m \): \[ m = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} \] Thus, - \( m_1 = \frac{12}{2} = 6 \) - \( m_2 = \frac{-4}{2} = -2 \) So, the complementary solution is: \[ y_c = C_1 x^6 + C_2 x^{-2} \] --- ## **Step 3: Find a Particular Solution** We need a particular solution for: \[ x^2 y'' - 3x y' - 12y = x^6 + \ln x \] **(a) For \( x^6 \) term:** Try \( y_{p1} = A x^6 \): \[ y_{p1}' = 6A x^5 \\ y_{p1}'' = 30A x^4 \] Plug into the equation: \[ x^2 [30A x^4] - 3x [6A x^5] - 12[A x^6] = x^6 \\ 30A x^6 - 18A x^6 - 12A x^6 = x^6 \\ (30A - 18A - 12A)x^6 = x^6 \\ 0 \cdot x^6 = x^6 \] So, \( A \) drops out (because \( x^6 \) is a solution of the homogeneous equation). Try \( y_{p1} = B x^6 \ln x \): \[ y_{p1}' = B (6x^5 \ln x + x^5) \\ y_{p1}'' = B [6 \cdot 5 x^4 \ln x + 6x^4 + 5x^4] = B [30x^4 \ln x + 11x^4] \] Now: \[ x^2 y_{p1}'' = B x^2 [30x^4 \ln x + 11x^4] = B [30x^6 \ln x + 11x^6] \\ -3x y_{p1}' = -3B x [6x^5 \ln x + x^5] = -3B [6x^6 \ln x + x^6] = -18B x^6 \ln x - 3B x^6 \\ -12y_{p1} = -12B x^6 \ln x \\ \] Add together: \[ x^2 y_{p1}'' - 3x y_{p1}' - 12y_{p1} = [30B x^6 \ln x + 11B x^6] + [-18B x^6 \ln x - 3B x^6] + [-12B x^6 \ln x] \\ = (30B - 18B - 12B)x^6 \ln x + (11B - 3B)x^6 \\ = 0 \cdot x^6 \ln x + 8B x^6 \] Set equal to \( x^6 \): \[ 8B x^6 = x^6 \implies B = \frac{1}{8} \] So, one particular solution is: \[ y_{p1} = \frac{1}{8} x^6 \ln x \] --- **(b) For \( \ln x \) term:** Try \( y_{p2} = a (\ln x) + b \): Now, calculate derivatives: \[ y_{p2} = a \ln x + b \\ y_{p2}' = \frac{a}{x} \\ y_{p2}'' = -\frac{a}{x^2} \] Substitute into the equation: \[ x^2 \left(-\frac{a}{x^2}\right) - 3x \left(\frac{a}{x}\right) - 12(a \ln x + b) = \ln x \\ -a - 3a - 12a \ln x - 12b = \ln x \\ -4a - 12a \ln x - 12b = \ln x \] Match coefficients with the right-hand side: For \( \ln x \): \[ -12a = 1 \implies a = -\frac{1}{12} \] For constants: \[ -4a - 12b = 0 \implies -4(-1/12) - 12b = 0 \implies \frac{1}{3} - 12b = 0 \implies b = \frac{1}{36} \] So, the second particular solution is: \[ y_{p2} = -\frac{1}{12} \ln x + \frac{1}{36} \] --- ## **Step 4: General Solution** Combine all: \[ \boxed{ y(x) = C_1 x^6 + C_2 x^{-2} + \frac{1}{8} x^6 \ln x - \frac{1}{12} \ln x + \frac{1}{36} } \] --- ### **Final Answer** \[ \boxed{ y(x) = C_1 x^6 + C_2 x^{-2} + \frac{1}{8} x^6 \ln x - \frac{1}{12} \ln x + \frac{1}{36} } \]

Let's carefully solve the given differential equation step-by-step, providing detailed calculations and explanations. --- ### **Given Differential Equation:** \[ x^2 y'' - 3x y' - 12 y = e^{6 \ln x} + \ln x \] --- ### **Step 1: Simplify the Right-Hand Side** Note that: \[ e^{6 \ln x} = (e^{\ln x})^6 = x^6 \] So, the equation becomes: \[ x^2 y'' - 3x y' - 12 y = x^6 + \ln x \] --- ### **Step 2: Solve the Homogeneous Equation** Homogeneous part: \[ x^2 y'' - 3x y' - 12 y = 0 \] This is an **Euler-Cauchy** (or equidimensional) differential equation. We assume a solution of the form: \[ y = x^m \] then: \[ y' = m x^{m-1} \] \[ y'' = m (m - 1) x^{m - 2} \] Substitute into the homogeneous equation: \[ x^2 [m (m - 1) x^{m-2}] - 3x [m x^{m-1}] - 12 x^{m} = 0 \] Simplify each term: \[ m (m - 1) x^{m} - 3m x^{m} - 12 x^{m} = 0 \] Factor out \( x^{m} \): \[ [x^{m}] \left[ m (m - 1) - 3m - 12 \right] = 0 \] Characteristic **quadratic equation**: \[ m (m - 1) - 3m - 12 = 0 \] \[ m^2 - m - 3m - 12 = 0 \] \[ m^2 - 4m - 12 = 0 \] Solve for \( m \): \[ m = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-12)}}{2} = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} \] \[ m = \frac{4 \pm 8}{2} \] Thus, - \( m_1 = \frac{4 + 8}{2} = \frac{12}{2} = 6 \) - \( m_2 = \frac{4 - 8}{2} = \frac{-4}{2} = -2 \) **Homogeneous solution:** \[ \boxed{ y_h = C_1 x^{6} + C_2 x^{-2} } \] --- ### **Step 3: Find a Particular Solution** Our nonhomogeneous term is: \[ x^6 + \ln x \] We will find particular solutions for each separately and then sum. --- ### **(a) Particular solution for \( x^6 \):** Since \( x^6 \) is a solution to the homogeneous equation (it's associated with \( m=6 \)), we need to try a solution **of the form**: \[ y_{p1} = A x^6 \ln x \] (Since the standard power solution overlaps with the homogeneous solution, multiplying by \( \ln x \) is a standard method to find a particular solution in such cases.) --- #### Compute derivatives: \[ y_{p1} = A x^6 \ln x \] \[ y_{p1}' = A \left( 6 x^{5} \ln x + x^{5} \right) = A x^{5} (6 \ln x + 1) \] \[ y_{p1}'' = A \frac{d}{dx} \left( x^{5} (6 \ln x + 1) \right) \] Apply product rule: \[ y_{p1}'' = A \left[ 5 x^{4} (6 \ln x + 1) + x^{5} \left( \frac{6}{x} \right) \right] = A \left[ 5 x^{4} (6 \ln x + 1) + 6 x^{4} \right] \] Simplify: \[ y_{p1}'' = A x^{4} \left[ 5 (6 \ln x + 1) + 6 \right] = A x^{4} \left( 30 \ln x + 5 + 6 \right) = A x^{4} (30 \ln x + 11) \] --- ### **Substitute into the differential equation:** \[ x^2 y_{p1}'' - 3x y_{p1}' - 12 y_{p1} = x^6 \] Calculate each term: 1. \( x^2 y_{p1}'' = x^2 \times A x^{4} (30 \ln x + 11) = A x^{6} (30 \ln x + 11) \) 2. \( -3x y_{p1}' = -3x \times A x^{5} (6 \ln x + 1) = -3A x^{6} (6 \ln x + 1) \) 3. \( -12 y_{p1} = -12 \times A x^{6} \ln x = -12A x^{6} \ln x \) Add all: \[ A x^{6} (30 \ln x + 11) - 3A x^{6} (6 \ln x + 1) - 12A x^{6} \ln x = x^{6} \] Factor \( A x^{6} \): \[ A x^{6} \left[ (30 \ln x + 11) - 3 (6 \ln x + 1) - 12 \ln x \right] = x^{6} \] Simplify inside brackets: \[ 30 \ln x + 11 - 18 \ln x - 3 - 12 \ln x = x^{6} \] Combine like terms: \[ (30 \ln x - 18 \ln x - 12 \ln x) + (11 - 3) = x^{6} \] \[ (0) + 8 = x^{6} \] So: \[ A x^{6} \times 8 = x^{6} \implies 8A x^{6} = x^{6} \] Divide both sides by \( x^{6} \): \[ 8A = 1 \implies A = \frac{1}{8} \] --- ### **Particular solution for \( x^6 \):** \[ \boxed{ y_{p1} = \frac{1}{8} x^{6} \ln x } \] --- ### **(b) Particular solution for \( \ln x \):** Since \( \ln x \) is not a solution of the homogeneous equation, try: \[ y_{p2} = a \ln x + b \] Calculate derivatives: \[ y_{p2}' = \frac{a}{x} \] \[ y_{p2}'' = -\frac{a}{x^{2}} \] Substitute into the differential equation: \[ x^{2} y_{p2}'' - 3x y_{p2}' - 12 y_{p2} = \ln x \] Compute each term: 1. \( x^{2} y_{p2}'' = x^{2} \times -\frac{a}{x^{2}} = -a \) 2. \( -3x y_{p2}' = -3x \times \frac{a}{x} = -3a \) 3. \( -12 y_{p2} = -12 (a \ln x + b) = -12a \ln x - 12b \) Sum: \[ (-a) - 3a - 12a \ln x - 12b = \ln x \] Combine constants: \[ (-a - 3a) - 12a \ln x - 12b = \ln x \] \[ -4a - 12a \ln x - 12b = \ln x \] Now, match coefficients: - Coefficient of \( \ln x \): \[ -12a = 1 \implies a = -\frac{1}{12} \] - Constant term: \[ -4a - 12b = 0 \] Substitute \( a = -\frac{1}{12} \): \[ -4 \times \left(-\frac{1}{12}\right) - 12b = 0 \\ \frac{4}{12} - 12b = 0 \\ \frac{1}{3} - 12b = 0 \implies 12b = \frac{1}{3} \implies b = \frac{1}{36} \] --- ### **Particular solution for \( \ln x \):** \[ \boxed{ y_{p2} = -\frac{1}{12} \ln x + \frac{1}{36} } \] --- ### **Step 4: Write the General Solution** Sum the homogeneous and particular solutions: \[ \boxed{ \boxed{ y(x) = C_1 x^6 + C_2 x^{-2} + \frac{1}{8} x^6 \ln x - \frac{1}{12} \ln x + \frac{1}{36} } } \] --- ### **Final Answer:** \[ \boxed{ \boxed{ \textbf{Solution:} \quad y(x) = C_1 x^6 + C_2 x^{-2} + \frac{1}{8} x^6 \ln x - \frac{1}{12} \ln x + \frac{1}{36} } } \] where \( C_1, C_2 \) are arbitrary constants. --- Let me know if you'd like further clarifications!