If y = aenx + be– nx, then prove that yn = n²y.

1.	If y = aenx + be– nx, then prove that yn = n²y.
Answer» Given y = ae^nx + be^{– nx} y¹ = ae^nx . n + be^{– nx} (–n) = ane^nx – bne^{– nx} y¹¹ = ane^nx . n – bne^{– nx}(–n) = an² e^nx + bn² e^{– nx} = n² (ae^nx + be^{– nx}) = n²y ∴ y¹¹ = n²y.

Answer»

Given y = ae^nx + be^{– nx}

y¹ = ae^nx . n + be^{– nx} (–n)

= ane^nx – bne^{– nx}

y¹¹ = ane^nx . n – bne^{– nx}(–n)

= an² e^nx + bn² e^{– nx}

= n² (ae^nx + be^{– nx})

= n²y

∴ y¹¹ = n²y.

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If y = aenx + be– nx, then prove that yn = n²y.

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