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Vector Algebra

Question
CBSEENMA12032906
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Find the differential equations from y = k esin–1 x   + 3.

Solution

The given equation is straight y space equals space ke to the power of sin to the power of negative 1 end exponent straight x end exponent plus 3                ...(1)
therefore space space space dy over dx equals space straight k space straight e to the power of sin to the power of negative 1 end exponent straight x end exponent. fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction space or space space dy over dx space equals space left parenthesis straight y minus 3 right parenthesis. space fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
or space space square root of 1 minus straight x squared end root space dy over dx equals straight y minus 3 comma space which is required differential equation.

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:
y' + y = ex

Determine order and degree (if defined) of differential equations:
y'' + (y')2 + 2 y = 0

Determine order and degree (if defined) of differential equations:
y″ + 2y′ + sin y = 0

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