Let's clarify the problem statement: Given: \[ x y'' - x y = e^x \] and it is given that \( y_1 = e^x \) is a solution to the complementary (homogeneous) equation. Let's find the general solution using **reduction of order**. --- ## 1. **Write the equation in standard form** Move all terms to one side and write in terms of \( y'' \): \[ x y'' - x y = e^x \implies y'' - y = \frac{e^x}{x} \] So, the equation is: \[ y'' - y = \frac{e^x}{x} \] --- ## 2. **Find the general solution to the homogeneous equation** The homogeneous equation is: \[ y'' - y = 0 \] The characteristic equation: \[ r^2 - 1 = 0 \implies r = \pm 1 \] So, the general solution is: \[ y_h = C_1 e^x + C_2 e^{-x} \] Given: \( y_1 = e^x \) is indeed a solution. --- ## 3. **Reduction of order** Since \( y_1 = e^x \) is a solution, let's look for a second solution of the form: \[ y_2(x) = v(x) e^x \] Plug \( y_2 \) into the homogeneous equation. Calculate derivatives: - \( y_2 = v e^x \) - \( y_2' = v' e^x + v e^x = (v' + v) e^x \) - \( y_2'' = (v'' + 2v' + v) e^x \) Plug into the homogeneous equation \( y'' - y = 0 \): \[ (v'' + 2v' + v) e^x - v e^x = 0 \] \[ (v'' + 2v') e^x = 0 \implies v'' + 2v' = 0 \] This is a first order equation in \( v' \): Let \( u = v' \): \[ u' + 2u = 0 \] \[ u' = -2u \] \[ \frac{du}{u} = -2 dx \] \[ \ln|u| = -2x + C \] \[ u = A e^{-2x} \] Now integrate \( u = v' \): \[ v' = A e^{-2x} \] \[ v = -A e^{-2x}/2 + B \] Therefore, the second solution is: \[ y_2 = v e^x = \left(-\frac{A}{2} e^{-2x} + B \right) e^x = -\frac{A}{2} e^{-x} + B e^x \] But recall, \( y_1 = e^x \), so the linearly independent solution is \( e^{-x} \). So the general solution to the homogeneous equation is: \[ y_h = C_1 e^x + C_2 e^{-x} \] --- ## 4. **Find a particular solution (variation of parameters)** Given the nonhomogeneous term \( \frac{e^x}{x} \), use variation of parameters. Let the particular solution be: \[ y_p = u_1(x) e^x + u_2(x) e^{-x} \] Where \( u_1' \) and \( u_2' \) are given by: \[ u_1' = -\frac{y_2(x) f(x)}{W(y_1, y_2)} \] \[ u_2' = \frac{y_1(x) f(x)}{W(y_1, y_2)} \] Where \( f(x) = \frac{e^x}{x} \), \( y_1 = e^x \), \( y_2 = e^{-x} \). Compute the Wronskian: \[ W = \begin{vmatrix} e^x & e^{-x} \\ e^x & -e^{-x} \end{vmatrix} = e^x(-e^{-x}) - e^{-x} e^x = -e^{x}e^{-x} - e^{-x}e^x = -2 \] Now plug in: \[ u_1' = -\frac{e^{-x} \cdot \frac{e^x}{x}}{-2} = \frac{e^{-x}e^x}{2x} = \frac{1}{2x} \] \[ u_2' = \frac{e^{x} \cdot \frac{e^x}{x}}{-2} = -\frac{e^{x}e^{x}}{2x} = -\frac{e^{2x}}{2x} \] Integrate: \[ u_1 = \int \frac{1}{2x} dx = \frac{1}{2} \ln|x| \] \[ u_2 = \int -\frac{e^{2x}}{2x} dx \] Let’s leave \( u_2 \) as is for now. So, \[ y_p = u_1(x) e^x + u_2(x) e^{-x} = \frac{1}{2} \ln|x| e^x - e^{-x} \int \frac{e^{2x}}{2x} dx \] --- ### The integral \( \int \frac{e^{2x}}{2x} dx \) does not have a simple elementary form. Let’s write it as it is. --- ## 5. **Final General Solution** \[ \boxed{ y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{2} e^x \ln|x| - \frac{1}{2} e^{-x} \int \frac{e^{2x}}{x} dx } \] Where \( \int \frac{e^{2x}}{x} dx \) is an indefinite integral that cannot be expressed in terms of elementary functions. --- ### **Summary of Steps:** 1. Write equation in standard form. 2. Solve homogeneous equation (\( C_1 e^x + C_2 e^{-x} \)). 3. Use reduction of order to find the second solution. 4. Use variation of parameters to find a particular solution. 5. Combine for the general solution. --- ## **Final Answer:** \[ \boxed{ y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{2} e^x \ln|x| - \frac{1}{2} e^{-x} \int \frac{e^{2x}}{x} dx } \] where \( C_1, C_2 \) are arbitrary constants.