In the following, verify that the given functions, (explicit or implicit) is a solution of the corresponding differential equation:
$straight x plus straight y space equals space tan to the power of negative 1 end exponent straight y space space space colon space space straight y squared straight y apostrophe space plus straight y squared plus 1 space equals space 0$

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Solution

straight x plus straight y space equals space tan to the power of negative 1 end exponent straight y

rightwards double arrow space space space space space 1 plus straight y apostrophe space equals space fraction numerator 1 over denominator 1 plus straight y squared end fraction. straight y apostrophe rightwards double arrow space 1 plus straight y squared plus straight y apostrophe plus straight y apostrophe straight y squared space equals straight y apostrophe rightwards double arrow space 1 plus straight y squared plus straight y apostrophe straight y squared space equals space 0 rightwards double arrow space straight y squared straight y apostrophe space plus space straight y squared plus 1 space equals space 0 therefore space space space space straight x plus straight y space equals space tan to the power of negative 1 end exponent straight y space space is space straight a space solution space of space straight y squared straight y apostrophe space plus space straight y squared plus 1 space equals space 0

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