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 x,

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

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

= Ame^mx + Bne^nx  ...(2)

Differentiating both sides again with respect to x,

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

= Am²e^mx + Bn²e^nx   ...(3)

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

= Am²e^mx + Bn²e^nx − (m + n) × (Ame^mx + Bne^nx) + mn (Ae^mx + Be^nx)  ...[Substituting the value (1), (2) and (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 × 0 + Be^nx × 0

= 0 = Right side