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Find the solution of the Diffrential Equation dy/dt = -| tan^5(y) |
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well, the absolute value looks a bit weird. break it up into two eqns , one when tan^5(y) >= 0 , so that will mean that |tan^5(y)| = tan^5(y) and one when tan^5(y)<0
so first case : dy/dt = - tan^5 (y) --> separate dy/ tan^5 (y) = -dt sec^5(y) dy = - dt --> now integrate break LHS into sec^4 and sec remember that sin^2 +cos^2 =1 , so sec^2 = 1 + tan^2
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