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Let's begin by solving the given second-order linear differential equation for the particular solution of the form \( y = e^{mx} \). The general differential equation is: \[ \frac{d^2y}{dx^2} + a(x)\frac{dy}{dx} + b(x)y = 0 \] We are asked to show that \( e^{mx} \) is a particular solution when \( m^2 + a(m) + b(m) = 0 \), where \( a(x) \) and \( b(x) \) are functions of \( x \). ### Step 1: Substituting \( y = e^{mx} \) into the differential equation For the equation \( y = e^{mx} \), the first and second derivatives are: \[ \frac{dy}{dx} = m e^{mx}, \quad \frac{d^2y}{dx^2} = m^2 e^{mx} \] Substituting these into the differential equation: \[ m^2 e^{mx} + a(x) m e^{mx} + b(x) e^{mx} = 0 \] We can factor out \( e^{mx} \), which is never zero: \[ e^{mx}(m^2 + a(x) m + b(x)) ___ __________ _______ _______ ____ ____ ________ ___ __________ ______.
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