\( e^{-2x}\times y=\int _{}^{}xe^{-2x} d \)

To solve the equation \( e^{-2x} \times y = \int xe^{-2x} \, dx \), we first need to compute the integral on the right-hand side. To evaluate the integral \( \int xe^{-2x} \, dx \), we can use integration by parts. We set: - \( u = x \) (which means \( du = dx \)) - \( dv = e^{-2x} \, dx \) (which means \( v = -\frac{1}{2} e^{-2x} \)) Now, applying integration by parts: \[ \int u \, dv = uv - \int v \, du \] Substituting our choices: \[ \int xe^{-2x} \, dx = x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) \, dx \] This simplifies to: \[ -\frac{1}{2} x e^{-2x} + \frac{1}{2} \int e^{-2x} \, dx \] Now, we need to compute \( \int e^{-2x} \, dx \): \[ \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} \] Substituting this back into our equation gives: \[ \int xe^{-2x} \, dx = -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} + C \] where \( C \) is the constant of integration. Now, we can substitute this result back into the original equation: \[ e^{-2x} y = -\frac{1}{2} x e^{-2x} - \frac{1}{4} e^{-2x} + C \] To isolate \( y \), we multiply both sides by \( e^{2x} \): \[ y = -\frac{1}{2} x - \frac{1}{4} + Ce^{2x} \] Thus, the solution for \( y \) is: \[ y = -\frac{1}{2} x - \frac{1}{4} + Ce^{2x} \] where \( C \) is an arbitrary constant.