If show that

It is given that, $y=A e^{m x}+B e^{n x}$

Then,

$\frac{d y}{d x}=A \cdot \frac{d}{d x}\left(e^{m x}\right)+B \cdot \frac{d}{d x}\left(e^{m x}\right)=A \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B \cdot e^{n x} \cdot \frac{d}{d x}(n x)=A m e^{m x}+B n e^{n x}$

$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(A m e^{m x}+B n e^{n x}\right)=A m \cdot \frac{d}{d x}\left(e^{m x}\right)+B n \cdot \frac{d}{d x}\left(e^{n x}\right)$

$=A m \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B n \cdot e^{n x} \cdot \frac{d}{d x}(n x)=A m^{2} e^{m x}+B n^{2} e^{n x}$

$\therefore \frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y$

$=A m^{2} e^{m x}+B n^{2} e^{n x}-(m+n) \cdot\left(A m e^{m x}+B n e^{n x}\right)+m n\left(A e^{m x}+B e^{n x}\right)$

$=A m^{2} e^{m x}+B n^{2} e^{n x}-A m^{2} e^{m x}-B m n e^{n x}-A m n e^{m x}-B n^{2} e^{m x}+A m n e^{m x}+B m n e^{n x}$

$=0$

Hence, proved.