Let's solve the problem step-by-step in English: --- ### **Step 1: Show transformation to constant coefficients** Given the differential equation: \[ x^2 y'' + \alpha x y' + \beta y = 0 \] with the change of variable \( z = \ln x \). #### **Compute derivatives:** Let \( y(x) = Y(z) \), where \( z = \ln x \implies x = e^z \). - \(\frac{dy}{dx} = \frac{dY}{dz} \frac{dz}{dx} = Y'(z) \cdot \frac{1}{x} \) - \(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(Y'(z) \cdot \frac{1}{x}\right)\) Compute \(\frac{d^2y}{dx^2}\): \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(Y'(z) \cdot \frac{1}{x}\right) \] Using the product rule: \[ = Y''(z) \left(\frac{dz}{dx}\right) \cdot \frac{1}{x} + Y'(z) \cdot \frac{d}{dx}\left(\frac{1}{x}\right) \] But \(\frac{dz}{dx} = \frac{1}{x}\), so: \[ = Y''(z) \cdot \frac{1}{x} \cdot \frac{1}{x} + Y'(z) \cdot \left(-\frac{1}{x^2}\right) = Y''(z) \cdot \frac{1}{x^2} - Y'(z) \cdot \frac{1}{x^2} \] So, \[ \frac{d^2y}{dx^2} = \frac{1}{x^2} \left(Y''(z) - Y'(z)\right) \] #### **Substitute into original equation:** \[ x^2 \frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0 \] \[ x^2 \left(\frac{1}{x^2} (Y'' - Y')\right) + \alpha x \left(\frac{1}{x} Y'\right) + \beta Y = 0 \] \[ (Y'' - Y') + \alpha Y' + \beta Y = 0 \] \[ Y'' + (\alpha - 1) Y' + \beta Y = 0 \] **This is a constant coefficient ODE in terms of \(z\) (or \(Y\)).** --- ### **Step 2: Solve the second equation** Given: \[ x^2 y'' + 2 x y' + 4 y = 4 \sin(\ln x) + e^{2 \ln x} \] #### **Change of variable: \(z = \ln x\), \(y(x) = Y(z)\)** From above: \[ x^2 y'' + 2 x y' + 4 y = Y'' + (2-1)Y' + 4Y = Y'' + Y' + 4Y \] And: - \(\sin(\ln x) = \sin(z)\) - \(e^{2 \ln x} = (e^{\ln x})^2 = x^2\) So, the equation becomes: \[ Y'' + Y' + 4Y = 4 \sin(z) + x^2 \] But recall \(x^2 = e^{2z}\): \[ Y'' + Y' + 4Y = 4 \sin(z) + e^{2z} \] --- ### **Step 3: Solve the ODE with constant coefficients** The equation: \[ Y'' + Y' + 4Y = 4\sin(z) + e^{2z} \] #### **Solve the homogeneous equation:** \[ Y'' + Y' + 4Y = 0 \] Characteristic equation: \[ r^2 + r + 4 = 0 \implies r = \frac{-1 \pm \sqrt{1 - 16}}{2} = \frac{-1 \pm i\sqrt{15}}{2} \] So, \[ Y_h(z) = e^{-z/2}\left(C_1 \cos\left(\frac{\sqrt{15}}{2}z\right) + C_2 \sin\left(\frac{\sqrt{15}}{2}z\right)\right) \] #### **Particular solution for \(4 \sin(z)\):** Try \(Y_p^{(1)} = A \sin(z) + B \cos(z)\): \[ Y_p^{(1)\prime} = A \cos(z) - B \sin(z) \] \[ Y_p^{(1)\prime\prime} = -A \sin(z) - B \cos(z) \] Substitute: \[ (-A \sin z - B \cos z) + (A \cos z - B \sin z) + 4(A \sin z + B \cos z) = 4 \sin z \] Group: \[ (-A + 4A) \sin z + (A) \cos z + (-B + 4B) \cos z + (-B) \sin z = 4 \sin z \] \[ (3A - B)\sin z + (A + 3B)\cos z = 4\sin z \] Set equal to \(4\sin z + 0\cos z\): \[ 3A - B = 4 \\ A + 3B = 0 \] Solving: From the second equation: \(A = -3B\) Substitute into the first: \[ 3(-3B) - B = 4 \implies -9B - B = 4 \implies -10B = 4 \implies B = -\frac{2}{5} \] \[ A = -3B = \frac{6}{5} \] So, \[ Y_p^{(1)} = \frac{6}{5}\sin z - \frac{2}{5}\cos z \] #### **Particular solution for \(e^{2z}\):** Try \(Y_p^{(2)} = C e^{2z}\): \[ Y_p^{(2)\prime} = 2C e^{2z}, \quad Y_p^{(2)\prime\prime} = 4C e^{2z} \] Plug into the ODE: \[ 4C e^{2z} + 2C e^{2z} + 4C e^{2z} = e^{2z} \] \[ (4C + 2C + 4C) e^{2z} = e^{2z} \implies 10C = 1 \implies C = \frac{1}{10} \] So, \[ Y_p^{(2)} = \frac{1}{10} e^{2z} \] --- ### **Step 4: General solution in \(z\):** \[ Y(z) = e^{-z/2} \left[C_1 \cos\left(\frac{\sqrt{15}}{2}z\right) + C_2 \sin\left(\frac{\sqrt{15}}{2}z\right)\right] + \frac{6}{5}\sin z - \frac{2}{5}\cos z + \frac{1}{10}e^{2z} \] --- ### **Step 5: Write the solution in terms of \(x\):** Recall \(z = \ln x\): - \(e^{-z/2} = x^{-1/2}\) - \(\sin(z) = \sin(\ln x)\) - \(\cos(z) = \cos(\ln x)\) - \(e^{2z} = x^2\) So: \[ y(x) = x^{-1/2} \left[C_1 \cos\left(\frac{\sqrt{15}}{2} \ln x\right) + C_2 \sin\left(\frac{\sqrt{15}}{2} \ln x\right)\right] + \frac{6}{5} \sin(\ln x) - \frac{2}{5} \cos(\ln x) + \frac{1}{10} x^2 \] --- ## **Final Answer:** ### **1. The equation \(x^2 y'' + \alpha x y' + \beta y = 0\) can be transformed into a constant-coefficient equation using \(z = \ln x\):** \[ Y'' + (\alpha - 1) Y' + \beta Y = 0 \] ### **2. Solution for \(x^2 y'' + 2 x y' + 4 y = 4 \sin(\ln x) + e^{2 \ln x}\):** \[ \boxed{ y(x) = x^{-1/2} \left[C_1 \cos\left(\frac{\sqrt{15}}{2} \ln x\right) + C_2 \sin\left(\frac{\sqrt{15}}{2} \ln x\right)\right] + \frac{6}{5} \sin(\ln x) - \frac{2}{5} \cos(\ln x) + \frac{1}{10} x^2 } \]