Question

If y = Ae^mx + Be^nx, show that `(d^2y)/dx^2  - (m+ n) (dy)/dx + mny = 0`

Answer 1

Given, y = Ae^mx + be^nx   ...(1)

Differentiating both sides with respect to ,

`dy/dx = A  d/dx  e^(mx) + B  d/dx  e^(nx) `

`= A  e^(mx) d/dx  (mx) + B  e^(nx) d/dx  (nx)`

= A · me^mx + B · ne^nx     ...(2)

Differentiating both sides again with respect to x,

`(d^2 y)/dx^2 = Am  d/dx  e^(mx) + Bn  d/dx  e^(nx)`

= Am² e^mx + Bn² e^nx          ...(3)

left side `(d^2 y)/dx^2 - (m + n) dy/dx + mn y `

= Am² e^mx + Bn² e^nx - (m + n) x (Ame^mx + Bne^nx) + mn (Ae^mx + Be^nx)

...[Substituting the value of y from equation (1), `dy/dx` from equation (2) and `(d^2 y)/dx^2` from equation (3)]

= Ae^mx [m² - m(m + n) + mn] + Be^nx [n² - n (m + n) + mn]

= Ae^mx [m² - m² - mn + mn] + Be^nx [n² - mn - n² + mn]

= Ae^mx x 0 + Be^nx x 0 = 0 = right side

Answer 2

Let y = Ae^mx +  Be^nx                ....(1)

Differentiating (1) w.r.t x we get

`dy/dx = Ae^(mx). m+ Be^(nx).n = Ame^(mx) + Bn e^(nx)`       ....(2)

Differentiating (2) w.r.t.x, we get,

`(d^2y)/dx^2 = Ame^(mx). m + Bn e^(nx).n`

`= Am^2e^(mx) + Bn^2e^(nx)`

Now,

`(d^2y)/dx^2 - (m + n) dy/dx + mny `

`= Am^2e^(mx) + Bn^2e^(nx) - [(m + n) Ame^(mx) + Bn e^(nx) ] + mn (Ae^(mx) + Be^(nx))`      .....[from (1), (2) and  (3)]

`= Am^2e^(mx) - Bn^2e^(nx) - Am^2e^(mx) - Bmn e^(nx) - Amn e^(mx) - Bn^2e^(nx) + Amn e^(mx) + Bmn e^(nx) = 0`