Question

If
$\rm \implies \: y = A {e}^{mx} + B {e}^{nx}$
Prove that
$\rm \implies \: \dfrac{d {}^{2}y }{d {x}^{2} } - (m + n) \dfrac{dy}{dx} + mny = 0$
​

Answer 1

Step-by-step explanation:

We have,

$y = a {e}^{mx} + b {e}^{nx}$

$\implies \frac{dy}{dx} = am {e}^{mx} + bn {e}^{nx}$

$\implies \: (m + n) \frac{dy}{dx} = (m + n)(am {e}^{mx} + bn {e}^{nx} )$

$\implies(m + n) \frac{dy}{dx} = a {m}^{2} {e}^{mx} + mn( a{e}^{mx} + b{e}^{nx} ) + b {n}^{2} {e}^{nx}$

$\implies(m + n) \frac{dy}{dx} = a {m}^{2} {e}^{mx} + b {n}^{2} {e}^{nx} + mny$

Now,

$\frac{ {d}^{2}y }{d {x}^{2} } = a {m}^{2} {e}^{mx} + b {n}^{2} {e}^{nx}$

Now,

$\frac{ {d}^{2}y }{d {x}^{2} } - (m + n) \frac{dy}{dx} + mny$

$= a {m}^{2} {e}^{mx} + b {n}^{2} {e}^{nx} - a {m}^{2} {e}^{mx} - b {n}^{2} {e}^{nx} - mny + mny$

$= 0$

Answer 2

Given :

$\sf y = A {e}^{mx} + B {e}^{nx}$

To prove :

$\sf \implies \: \dfrac{d {}^{2}y }{d {x}^{2} } - (m + n) \dfrac{dy}{dx} + mny = 0$

Proof :

$\sf y = A {e}^{mx} + B {e}^{nx}$

Now first we will find out dy/dx

$\sf \implies \dfrac{dy}{dx} = \dfrac{d \bigg(A {e}^{mx} + B {e}^{nx} \bigg)}{dx}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{d (A {e}^{mx} )}{dx} + \dfrac{d(B {e}^{nx})}{dx}$

By differentiating using chain rule we get :-

$\sf \implies \dfrac{dy}{dx} = Am \: {e}^{mx} +Bn \: {e}^{nx}$

Now multiply m+n both sides.

$\sf \implies (m + n) \dfrac{dy}{dx} = (m + n) (Am \: {e}^{mx} +Bn \: {e}^{nx})$

$\sf \implies (m + n) \dfrac{dy}{dx} = A {m}^{2} \: {e}^{mx} +Bn m \: {e}^{nx} + A {m}n \: {e}^{mx}+B {n}^{2} \: {e}^{nx}$

taking out mn common :-

$\sf \implies (m + n) \dfrac{dy}{dx} = A {m}^{2} \: {e}^{mx} +nm(B\: {e}^{nx} + A \: {e}^{mx})+B {n}^{2} \: {e}^{nx}$

$\sf B\: {e}^{nx} + A \: {e}^{mx} = y$

$\sf \implies (m + n) \dfrac{dy}{dx} = A {m}^{2} \: {e}^{mx} +nmy+B {n}^{2} \: {e}^{nx}........(1)$

$\rm \: Now \: we \: will \: find \: out \: : \dfrac{d {}^{2}y }{d {x}^{2} }$

$\sf \implies \dfrac{d {}^{2}y }{d {x}^{2} } = \dfrac{d \bigg(Am \: {e}^{mx} +Bn \: {e}^{nx} \bigg)}{dx}$

$\sf \implies \dfrac{d {}^{2}y }{d {x}^{2} } = \dfrac{d \bigg(Am \: {e}^{mx}\bigg)}{dx} + \dfrac{d \bigg(Bn \: {e}^{nx} \bigg)}{dx}$

By differentiating using chain rule we get :-

$\sf \implies \dfrac{d {}^{2}y }{d {x}^{2} } = A {m}^{2} {e}^{mx} + B {n}^{2} \: {e}^{nx} .......(2)$

Now put the value of (1) and (2) in :-

$\sf \implies \: \dfrac{d {}^{2}y }{d {x}^{2} } - (m + n) \dfrac{dy}{dx} + mny$

$\sf \implies \: A {m}^{2} {e}^{mx} + B {n}^{2} \: {e}^{nx}- ( A {m}^{2} \: {e}^{mx} +nmy+B {n}^{2} \: {e}^{nx})+ mny$

$\sf \implies \: A {m}^{2} {e}^{mx} + B {n}^{2} \: {e}^{nx}- A {m}^{2} \: {e}^{mx} - nmy - B {n}^{2} \: {e}^{nx}+ mny$

$\sf \implies \: A {m}^{2} {e}^{mx} - A {m}^{2} \: {e}^{mx}+ B {n}^{2} \: {e}^{nx} - B {n}^{2} \: {e}^{nx} + mny - mny = 0$

Hence proved