Mathematics - Functions Question with Solution | TestHub

Let $f(x)=e^{mx}$ be a trial solution. Then $f(x+2)=e^{m(x+2)}=e^{mx}\cdot e^{2m}$, and $f(x+1)=e^{mx}\cdot e^m$. Therefore, $f(x+2)-5f(x+1)+6f(x)=0$.

$$e^{mx}\cdot e^{2m}-5e^{mx}\cdot e^m+6e^{mx}=0$$

$$e^{mx}(e^{2m}-5e^m+6)=0$$

Since $e^{mx}\ne 0$, we have:

$$e^{2m}-5e^m+6=0$$

Let $e^m=u$. Then from the above equation:

$$u^2-5u+6=0$$

$$(u-2)(u-3)=0$$

So, $u=2$ or $u=3$.

Therefore, $e^m=2$ or $e^m=3$.

This implies $e^{mx}=2^x$ or $e^{mx}=3^x$.

Hence, $f(x)=A\cdot 2^x+B\cdot 3^x$.