If y = eax + e-ax, Show that y2 – a2y = 0

1.	If y = eax + e-ax, Show that y2 – a2y = 0
Answer» y = e^ax + e^-ax y₁ = ae^ax – ae^-ax y₂ = a²e^ax + a²e^-ax = a² (e^ax + e^-ax) = a²y ∴ y² – a²y = 0.

Answer»

y = e^ax + e^-ax

y₁ = ae^ax – ae^-ax

y₂ = a²e^ax + a²e^-ax

= a² (e^ax + e^-ax) = a²y

∴ y² – a²y = 0.

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