Question

If $y=\left(A+Bx\right)e^{mx}+{\left(m-1\right)}^{-2}{e}^x$ 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. $e^{\left(1-m\right)x}$

Answer 1

The correct option is A $e^x$

given,

$y=(A+Bx)e^{mx}+(m-1{)}^{-2}{e}^x$

A,B and m are constant

differentiating on both sides w.r.t x

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

$\frac{dy}{dx}=B.e^{mx}+m\left(\right(A+Bx)e^{mx}+\frac{e^x}{(m-1{)}^2})-\frac{m{e}^x}{(m-1{)}^2}+\frac{m{e}^x}{(m-1{)}^2}$

$\frac{dy}{dx}=B.e^{mx}+my-\frac{e^x}{(m-1)}$

again differentiating on both sides

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

$=m\frac{dy}{dx}+B.m{e}^{mx}-\frac{m{e}^x}{m-1}+m^2+\frac{m{e}^x}{m-1}-m^2-\frac{e^x}{m-1}$

$=m\frac{dy}{dx}+m(B{e}^{mx}+my-\frac{e^x}{m-1})+e^x\left(\frac{m-1}{m-1}\right)-m^{2}y$

$\therefore \frac{d^{2}y}{d{x}^2}=m\frac{dy}{dx}+m\frac{dy}{dx}+e^x-m^{2}y$

$\therefore \frac{d^{2}y}{d{x}^2}-2m\frac{dy}{dx}+-m^{2}y=e^x$