Solution:
We need to Prove: $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$
$\frac{d y}{d x}=\mathrm{A} e^{m x} \frac{d}{d x}(m x)+\mathrm{B} e^{n x} \frac{d}{d x}(n x)\left[\because \frac{d}{d x} e^{f(x)}=e^{f(x)} \frac{d}{d x} f(x)\right]$
$\frac{d y}{d x}=\mathrm{A} m e^{m \mathrm{c}}+\mathrm{B} n e^{\pi x} \ldots(2)$
Find the derivate of equation (2)
$\frac{d^{2} y}{d x^{2}}=\mathrm{A} m e^{m \mathrm{c}} m+\mathrm{B} n e^{m x} \cdot n$
So now,
L.H.S. $=\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y$
(Using the eqs. (1), (2) and (3))
$=\mathrm{A} m^{2} e^{m x}+\mathrm{B} n^{2} e^{\prime x}-(m+n) \mathrm{A} m e^{\pi x}+\mathrm{B} n e^{m x}+m n\left(\mathrm{~A} e^{m x}+\mathrm{B} e^{m x}\right)$
$=A m^{2} e^{m x}+B n^{2} e^{m}-A m^{2} e^{m x}-B m n e^{m x}+A m n e^{m x}-B n^{2} e^{z x}+A m n e^{m x}+B m n e^{m}$
$=0$
$=$ R.H.S.
As a result hence proved.