Question

If y = Ae^mx + Be^nx, show that y₂ – (m + n)y₁ + mny = 0.

Answer 1

y = Ae^mx + Be^nx 
Differentiating w.r.t. x, we get

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

= `"Ae"^(mx)."d"/"dx"(mx) + "Be"^(nx)."d"/"dx"(nx)`

= Ae^mx . m + Be^nx . n
= y₁ = mAe^mx + nBe^nx                    ...(1)
Differentiating again w.r.t. x, we get

y₂ = `m"A""d"/"dx"(e^(mx)) + n"B""d"/"dx"(e^(nx))`

 = `m"Ae"^(mx)."d"/"dx"(mx) + n"Be"^(nx)."d"/"dx"(nx)`

= mAe^mx . m + nBe^nx . n

∴ y₂ = m²Ae^mx + n²Be^nx                   ...(2)
∴ y₂ – (m + n)y₁ + mny = (m²Ae^mx + n²Be^nx) – (m + n)(mAe^mx + nBe^nx) + mn(Ae^mx + Be^nx)                           ...[By (1), (2)]

= m²Ae^mx + n²Be^nx – m²Ae^mx – mnBe^mx – n²Be^nx + mnAe^mx + mnBe^nx

= 0
∴ y₂ – (m + n)y₁ + mny = 0