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If \( y = Ae^{mx} + Be^{nx}\), show that \( \large\frac{d^2y}{dx^2} -\normalsize ( m + n ) \large\frac{dy}{dx} +\normalsize mny = 0 \)

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Toolbox:
  • $y=f(x)$
  • $\large\frac{dy}{dx}$$=f'(x)$
  • $\large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big)$
  • $\large\frac{d}{dx}$$(e^x)=e^x$
Step 1:
$y=Ae^{\large mx}+Be^{\large nx}$
Differentiating with respect to $x$
$\large\frac{dy}{dx}$$=A.me^{\large mx}+B.ne^{\large nx}$
$\large\frac{d^2y}{dx^2}$$=A.m^2e^{\large mx}+B.n^2e^{\large nx}$
Step 2:
$\large\frac{d^2y}{dx^2}$$-(m+n)\large\frac{dy}{dx}$$+mny$
$\Rightarrow A.m^2e^{\large mx}+Bn^2e^{\large nx}-(m+n)(Ame^{\large mx}+Bne^{\large nx})+mn(Ae^{\large mx}+Be^{\large nx})$
$\quad\;=A.m^2e^{\large mx}+Bn^2e^{\large nx}-Am^2e^{\large mx}-Bmne^{\large nx}-Amne^{\large mx}-Bn^2e^{\large nx}+Amne^{\large mx}+Bmne^{\large nx}$
$\quad\;=0$
Hence proved.
answered May 13, 2013 by sreemathi.v
 
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