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2yy-y-2-1-8y-2-cos-2-x-

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Question Number 168325 by cortano1 last updated on 08/Apr/22
2yy′′−(y′)^2 −1= ((8y^2 )/(cos^2 x))
$$\:\:\:\:\:\:\mathrm{2}{yy}''−\left({y}'\right)^{\mathrm{2}} \:−\mathrm{1}=\:\frac{\mathrm{8}{y}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} {x}}\: \\ $$

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