Let's clarify your question: You have the differential equation: \[ x y'' - x y = e^x \] And you are told that \( y_1 = e^x \) is a solution to the complementary (homogeneous) equation. Let's solve it **step by step using reduction of order**. --- ### **Step 1: Write the Equation in Standard Form** First, rewrite the equation: \[ x y'' - x y = e^x \] Divide both sides by \( x \) (assuming \( x \neq 0 \)): \[ y'' - y = \frac{e^x}{x} \] **Explanation:** We divided by \( x \) to get the equation into standard linear ODE form, which is easier to analyze and apply known methods. --- ### **Step 2: Find the General Solution Using Reduction of Order** Given that \( y_1 = e^x \) is a solution to the homogeneous equation \( y'' - y = 0 \), let's find a second linearly independent solution \( y_2 \) using reduction of order. Assume \( y_2 = v(x) e^x \), where \( v(x) \) is to be determined. Compute derivatives: \[ y_2' = v' e^x + v e^x \] \[ y_2'' = v'' e^x + 2v' e^x + v e^x \] Plug into the homogeneous equation: \[ y_2'' - y_2 = [v'' e^x + 2v' e^x + v e^x] - [v e^x] = v'' e^x + 2v' e^x \] Set this equal to zero: \[ v'' e^x + 2v' e^x = 0 \implies v'' + 2v' = 0 \] Let \( u = v' \), then \( u' + 2u = 0 \) This is a separable first-order ODE: \[ \frac{du}{dx} + 2u = 0 \implies \frac{du}{u} = -2 dx \implies \ln|u| = -2x + C \implies u = v' = A e^{-2x} \] Integrate again: \[ v = \int A e^{-2x} dx = -\frac{A}{2} e^{-2x} + B \] So the general solution to the homogeneous equation is: \[ y_h = C_1 e^x + C_2 e^x \left( -\frac{1}{2} e^{-2x} \right ) = C_1 e^x + C_2 e^{-x} \] **Explanation:** We used reduction of order to find a second solution to the homogeneous equation, yielding the general solution to the homogeneous part. --- ### **Step 3: Find a Particular Solution** Now, find a particular solution to \[ y'' - y = \frac{e^x}{x} \] We use variation of parameters. The two linearly independent solutions to the homogeneous equation are \( y_1 = e^x \), \( y_2 = e^{-x} \). The particular solution is: \[ y_p = u_1(x) y_1(x) + u_2(x) y_2(x) \] Where: \[ u_1' = -\frac{y_2(x) f(x)}{W(y_1, y_2)}, \quad u_2' = \frac{y_1(x) f(x)}{W(y_1, y_2)} \] Calculate the Wronskian: \[ W = \begin{vmatrix} e^x & e^{-x} \\ e^x & -e^{-x} \end{vmatrix} = e^x \cdot (-e^{-x}) - e^{-x} \cdot e^x = -e^{x}e^{-x} - e^{-x}e^{x} = -1 - 1 = -2 \] Now, \[ u_1' = -\frac{e^{-x} \cdot \frac{e^x}{x}}{-2} = \frac{e^{-x} e^x}{2x} = \frac{1}{2x} \implies u_1 = \int \frac{1}{2x} dx = \frac{1}{2} \ln|x| \] \[ u_2' = \frac{e^{x} \cdot \frac{e^x}{x}}{-2} = -\frac{e^{2x}}{2x} \implies u_2 = -\frac{1}{2} \int \frac{e^{2x}}{x} dx \] Thus, \[ y_p = u_1(x) y_1(x) + u_2(x) y_2(x) = \frac{1}{2} \ln|x| \cdot e^x - \frac{1}{2} e^{-x} \int \frac{e^{2x}}{x} dx \] **Explanation:** We used variation of parameters to find a particular solution. The first term is straightforward, the second involves an integral that cannot be expressed in terms of elementary functions. --- ### **Final Answer** The **general solution** to the differential equation is: \[ \boxed{ y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{2} e^x \ln|x| - \frac{1}{2} e^{-x} \int \frac{e^{2x}}{x} dx } \] Where \( C_1, C_2 \) are arbitrary constants.

Let's clarify your question: You have the differential equation: \[ x y'' - x y = e^x \] and you're told that \( y_1 = e^x \) is a solution to the homogeneous (complementary) equation. --- ### **Step 1: Rewrite the differential equation in standard form** Divide both sides by \( x \) (assuming \( x \neq 0 \)): \[ y'' - y = \frac{e^x}{x} \] **Explanation:** Dividing by \( x \) transforms the equation into a standard linear second-order form: \[ y'' + p(x) y' + q(x) y = g(x) \] which here simplifies to: \[ y'' - y = \frac{e^x}{x} \] This form makes it easier to apply reduction of order and variation of parameters. --- ### **Step 2: Find the second linearly independent solution to the homogeneous equation** Homogeneous equation: \[ y'' - y = 0 \] Given \( y_1 = e^x \), find \( y_2 \) using reduction of order: Assume \( y_2 = v(x) e^x \). Compute derivatives: \[ y_2' = v' e^x + v e^x \] \[ y_2'' = v'' e^x + 2 v' e^x + v e^x \] Substitute into the homogeneous equation: \[ v'' e^x + 2 v' e^x + v e^x - v e^x = 0 \] which simplifies to: \[ v'' e^x + 2 v' e^x = 0 \] Divide through by \( e^x \): \[ v'' + 2 v' = 0 \] Let \( u = v' \), then: \[ u' + 2 u = 0 \] Solve for \( u \): \[ \frac{du}{dx} = -2 u \] \[ \Rightarrow \frac{du}{u} = -2 dx \] \[ \Rightarrow \ln|u| = -2x + C \] \[ \Rightarrow u = v' = A e^{-2x} \] Integrate: \[ v = \int A e^{-2x} dx = -\frac{A}{2} e^{-2x} + B \] Choose \( A = 1 \) for simplicity, then: \[ v = -\frac{1}{2} e^{-2x} + B \] Thus, the second solution: \[ y_2 = v e^x = \left( -\frac{1}{2} e^{-2x} \right) e^x = -\frac{1}{2} e^{-x} \] The general homogeneous solution: \[ y_h = C_1 e^x + C_2 e^{-x} \] --- ### **Final step: Find the particular solution using variation of parameters** Using \( y_1 = e^x \), \( y_2 = e^{-x} \), the Wronskian: \[ W = y_1 y_2' - y_1' y_2 = e^x (-e^{-x}) - e^x e^{-x} = -1 - 1 = -2 \] Compute: \[ u_1' = -\frac{y_2 g(x)}{W} = -\frac{e^{-x} \cdot \frac{e^x}{x}}{-2} = \frac{e^{-x} e^x / x}{2} = \frac{1}{2x} \] \[ u_2' = \frac{y_1 g(x)}{W} = \frac{e^x \cdot \frac{e^x}{x}}{-2} = - \frac{e^{2x}}{2x} \] Integrate: \[ u_1 = \frac{1}{2} \int \frac{1}{x} dx = \frac{1}{2} \ln|x| \] \[ u_2 = -\frac{1}{2} \int \frac{e^{2x}}{x} dx \] Therefore, the particular solution is: \[ y_p = u_1 y_1 + u_2 y_2 = \frac{1}{2} \ln|x| \cdot e^x - \frac{1}{2} e^{-x} \int \frac{e^{2x}}{x} dx \] --- ### **Final answer:** The **general solution** to the original differential equation is: \[ \boxed{ \mathbf{ y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{2} e^x \ln|x| - \frac{1}{2} e^{-x} \int \frac{e^{2x}}{x} dx } } \] where \( C_1, C_2 \) are arbitrary constants. **Note:** The integral \( \int \frac{e^{2x}}{x} dx \) does not have an elementary closed form, but this expression correctly represents the particular solution obtained via variation of parameters.