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If $y=A e^{\mathrm{mx}}+B e^{\mathrm{nx}}$, show that
$\frac{d^2 y}{d x^2}-(m+n) \frac{d y}{d x}+m n y=0$
$\frac{d^2 y}{d x^2}-(m+n) \frac{d y}{d x}+m n y=0$
Solution:
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Verified Answer
$\frac{d y}{d x}=A \cdot m e^{m x}+B \cdot n e^{n x}$
and $\quad \frac{d^2 y}{d x^2}=A \cdot m^2 e^{m x}+B \cdot n^2 e^{n x}$
L.H.S. $=\frac{d^2 y}{d x^2}-(m+n) \frac{d y}{d x}+m n y$
$=A \cdot m^2 e^{m x}+B \cdot n^2 e^{n x}-(m+n)$
$\quad-A \cdot m n e^{m x}-B \cdot n^2 e^{n x}+A \cdot m n e^{m x}+B \cdot m n e^{n x}$
=0=R.H.S
Hence, proved.
and $\quad \frac{d^2 y}{d x^2}=A \cdot m^2 e^{m x}+B \cdot n^2 e^{n x}$
L.H.S. $=\frac{d^2 y}{d x^2}-(m+n) \frac{d y}{d x}+m n y$
$=A \cdot m^2 e^{m x}+B \cdot n^2 e^{n x}-(m+n)$
$\quad-A \cdot m n e^{m x}-B \cdot n^2 e^{n x}+A \cdot m n e^{m x}+B \cdot m n e^{n x}$
=0=R.H.S
Hence, proved.
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