Question

If $y=(A+Bx)e^{mx}+(m-1{)}^{-2}{e}^x$ then $\frac{d^{2}y}{d{x}^2}-2m\frac{dy}{dx}+m^{2}y$ is equal to:

• A. $e^x$
• B. $e^{mx}$
• C. $e^{-mx}$
• D. $e^{(1-m)x}$

Answer 1

The correct option is A $e^x$

$y=(A+Bx)e^{mx}+(m-1{)}^{-2}{e}^x$
$m^{2}y=m^2(A+Bx)e^{mx}+m^2(m-1{)}^{-2}{e}^x$
$\frac{dy}{dx}=m(A+Bx)e^{mx}+B.e^{mx}+{(m-1)}^{-2}{e}^x$
$-2m\frac{dy}{dx}=-2m^2(A+Bx)e^{mx}-2mB.e^{mx}-2{m(m-1)}^{-2}{e}^x$
$\frac{d^{2}y}{d{x}^2}=m(m{e}^{mx}(A+Bx)+{Be}^{mx})+m.B{e}^{mx}+{(m-1)}^{-2}{e}^x$
$\frac{d^{2}y}{d{x}^2}=m^2{e}^{mx}(A+Bx)+mB{e}^{mx}+m.B{e}^{mx}+{(m-1)}^{-2}{e}^x$
$\frac{d^{2}y}{d{x}^2}=m^2{e}^{mx}(A+Bx)+2mB{e}^{mx}+{(m-1)}^{-2}{e}^x$