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