Question

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

Answer 1

It is given that,$y=Ae^{mx}+Be^{nx}$
Then,
$\frac{dy}{dx}=A.\frac{d}{dx}(e^{mx})+B.\frac{d}{dx}(e^{nx})$
$=A.e^{mx}.\frac{d}{dx}(mx)+B.e^{nx}.\frac{d}{dx}(nx)=Ame^{mx}+Bne^{nx}$
$\frac{d^2y}{dx^2}=\frac{d}{dx}(Ame^{mx}+Bne^{nx})$
$=Am.\frac{d}{dx}(e^{mx})+Bn\frac{d}{dx}(e^{nx})$
$=Am.e^{mx}.\frac{d}{dx}(mx)+Bn.e^{nx}.\frac{d}{dx}(nx)=Am^2e^{mx}+Bn^2e^{nx}$
$∴\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny$
$=Am^2e^{mx}+Bn^2e^{nx}-(m+n)(Ame^{mx}+Bne^{nx})+mn(Ae^{mx}+Be^{nx})$
$=Am^2e^{mx}+Bn^2e^{nx}-Am^2e^{mx}-Bmne^{nx}-Amne^{mx}-Bn^2e^{nx}+Amne^{mx}+Bmne^{nx}$
$=0$
Hence,proved